Solving Assignment Makes Integer from Pointer Without a Cast in C: Tips and Solutions
As a developer, encountering errors while coding is inevitable. One common error that C programmers come across is the "assignment makes integer from pointer without a cast" error. This error message can be frustrating and time-consuming to resolve, but with the right tips and solutions, it can be easily fixed.
Understanding the Error Message
Before we dive into the tips and solutions for fixing this error, let's first understand what it means. The "assignment makes integer from pointer without a cast" error occurs when a pointer is assigned to an integer without a proper type cast. This error message is often accompanied by a warning message that looks like this:
This warning message is telling the programmer that the code is trying to assign a pointer value to an integer variable without casting the pointer to the correct type.
Tips for Fixing the Error
Here are some tips to help you fix the "assignment makes integer from pointer without a cast" error:
Tip #1: Check Your Pointer Types
Make sure that the pointer you are trying to assign to an integer variable is of the correct data type. If the pointer is pointing to a different data type, you will need to cast it to the correct type before assigning it to the integer variable.
Tip #2: Use the Correct Syntax
When casting a pointer to a different data type, make sure to use the correct syntax. The syntax for casting a pointer to an integer is (int) pointer .
Tip #3: Use the Correct Assignment Operator
Make sure that you are using the correct assignment operator. The assignment operator for pointers is = while the assignment operator for integers is == .
Tip #4: Check Your Code for Errors
Double-check your code for errors. Sometimes, the "assignment makes integer from pointer without a cast" error can be caused by a syntax error or a missing semicolon.
Solutions for Fixing the Error
Now that you have some tips for fixing the "assignment makes integer from pointer without a cast" error, let's look at some solutions.
Solution #1: Cast the Pointer to the Correct Type
To fix this error, you need to cast the pointer to the correct type before assigning it to the integer variable. Here's an example:
In this example, the pointer is cast to an integer using the (int) syntax before it is assigned to the num variable.
Solution #2: Declare the Integer Variable as a Pointer
Another solution is to declare the integer variable as a pointer. Here's an example:
In this example, the num variable is declared as a pointer, and the ptr variable is assigned to it without casting.
Q1: What causes the "assignment makes integer from pointer without a cast" error?
A: This error occurs when a pointer is assigned to an integer variable without being cast to the correct data type.
Q2: How do I cast a pointer to an integer in C?
A: To cast a pointer to an integer in C, use the (int) syntax.
Q3: Why is my code still giving me the same error message even after I cast the pointer to the correct type?
A: Double-check your code for syntax errors and missing semicolons. Sometimes, these errors can cause the same error message to appear even after you have cast the pointer to the correct type.
Q4: Can I declare the integer variable as a pointer to fix this error?
A: Yes, you can declare the integer variable as a pointer to fix this error.
Q5: What is the correct assignment operator for pointers and integers in C?
A: The assignment operator for pointers is = while the assignment operator for integers is == .
The "assignment makes integer from pointer without a cast" error can be frustrating, but with the right tips and solutions, it can be easily fixed. By understanding the error message and following the tips and solutions provided in this guide, you can resolve this error and improve the functionality of your code.
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Assignment makes integer from pointer without a cast in c
Programming can be both rewarding and challenging. You work hard on your code, and just when it seems to be functioning perfectly, an error message pops up on your screen, leaving you frustrated and clueless about what went wrong. One common error that programmers encounter is the "Assignment makes integer from pointer without a cast" error in C.
This error occurs when you try to assign a value from a pointer variable to an integer variable without properly casting it. To fix this error, you need to make sure that you cast the pointer value to the appropriate data type before assigning it to an integer variable. In this article, we will dive deeper into the causes of this error and provide you with solutions to overcome it.
What makes this error occur?
I will present some cases that triggers that error to occur, and they are all have the same concept, so if you understanded why the failure happens, then you will figure out how to solve all the cases easily.
Case 1: Assignment of a pointer to an integer variable
In this simple code we have three variables, an integer pointer "ptr" , and two integers "n1" and "n2" . We assign 2 to "n1" , so far so good, then we assign the address of "n2" to "ptr" which is the suitable storing data type for a pointer, so no problems untill now, till we get to this line "n2 = ptr" when we try to assign "ptr" which is a memory address to "n2" that needs to store an integer data type because it's not a pointer.
Case 2: Returning a Pointer from a Function that Should Return an Integer
As you can see, it's another situation but it's the same idea which causes the compilation error. We are trying here to assign the return of getinteger function which is a pointer that conatains a memory address to the result variable that is an int type which needs an integer
Case 3: Misusing Array Names as Pointers
As we might already know, that the identifier (name) of the array is actually a pointer to the array first element memory address , so it's a pointer after all, and assigning a pointer type to int type causes the same compilation error.
The solutions
The key to avoiding the error is understanding that pointers and integers are different types of variables in C. Pointers hold memory addresses, while integers hold numeric values. We can use either casting , dereferencing the pointer or just redesign another solution for the problem we are working on that allows the two types to be the same. It all depending on the situation.
Let's try to solve the above cases:
Case 1: Solution: Deferencing the pointer
We need in this case to asssign an int type to "n2" not a pointer or memory address, so how do we get the value of the variable that the pointer "ptr" pointing to? We get it by deferencing the pointer , so the code after the fix will be like the following:
Case 2: Solution: Choosing the right data type
In this case we have two options, either we change the getinteger returning type to int or change the result variable type to a pointer . I will go with the latter option, because there are a lot of functions in the C standard library that returning a pointer, so what we can control is our variable that takes the function return. So the code after the fix will be like the following:
We here changed the result variable from normal int to an int pointer by adding "*" .
Case 3: Solution: Using the array subscript operator
In this case we can get the value of any number in the array by using the subscript opeartor ([]) on the array with the index number like: myarray[1] for the second element which is 2 . If we still remember that the array identifier is a pointer to the array first memory, then we can also get the value of the array first element by deferencing the array identifier like: *myarray which will get us 1 .
But let's solve the case by using the subscript opeartor which is the more obvious way. So the code will be like the following:
Now the number 1 is assigned to myint without any compilation erros.
The conclusion
In conclusion, the error "assignment to ‘int’ from ‘int *’ makes integer from pointer without a cast" arises in C programming when there is an attempt to assign a memory address (held by a pointer) directly to an integer variable. This is a type mismatch as pointers and integers are fundamentally different types of variables in C.
To avoid or correct this error, programmers need to ensure they are handling pointers and integers appropriately. If the intent is to assign the value pointed by a pointer to an integer, dereferencing should be used. If a function is meant to return an integer, it should not return a pointer. When dealing with arrays, remember that the array name behaves like a pointer to the first element, not an individual element of the array.
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[SOLVED] C - assigment makes integer from pointer without a cast warning
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Thread: [SOLVED] C - assigment makes integer from pointer without a cast warning
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I know it's a common error, and I've tried googling it, looked at a bunch of answers, but I still don't really get what to do in this situation.... Here's the relevant code: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char path[51]; const char* home = getenv( "HOME" ); strcpy( path, argv[1] ); path[1] = home; return 0; } -- there is more code in the blank lines, but the issue's not there (I'm fairly sure), so didn't see the point in writing out 100 odd lines of code. I've tried some stuff like trying to make a pointer to path[1], and make that = home, but haven't managed to make that work (although maybe that's just me doing it wrong as opposed to wrong idea?) Thanks in advance for any help
Re: C - assigment makes integer from pointer without a cast warning
path[1] is the second element of a char array, so it's a char. home is a char *, i.e. a pointer to a char. You get the warning because you try to assign a char* to a char. Note also the potential to overflow your buffer if the content of the argv[1] argument is very long. It's usually better to use strncpy.
Last edited by r-senior; March 10th, 2013 at 03:03 PM . Reason: argv[1] would overflow, not HOME. Corrected to avoid confusion.
Please create new threads for new questions. Please wrap code in code tags using the '#' button or enter it in your post like this: [code]...[/code].
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You can try something like this: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char *path; const char *home = getenv("HOME"); path = malloc(strlen(home) + 1); if (!path) { printf("Error\n"); return 0; } strcpy(path, home); printf("path = %s\n", path); // if you want to have argv[1] concatenated with path if (argc >= 2) { path = malloc(strlen(home) + strlen(argv[1]) + 1); strcpy(path, argv[1]); strcat(path, home); printf("%s\n", path); } // if you want an array of strings, each containing path, argv[1]... char **array; int i; array = malloc(argc * sizeof(char*)); array[0] = malloc(strlen(home) + 1); strcpy(array[0], home); printf("array[0] = %s\n", array[0]); for (i = 1; i < argc; i++) { array[i] = malloc(strlen(argv[i]) + 1); strcpy(array[i], argv[i]); printf("array[%d] = %s\n", i, array[i]); } // now array[i] will hold path and all the argv strings return 0; } Just as above, your path[51] is a string while path[1] is only a character, so you can't use strcpy for that.
Last edited by Christmas; March 10th, 2013 at 09:51 PM .
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Originally Posted by Christmas You can try something like this: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char *path; const char *home = getenv("HOME"); path = malloc(strlen(home) + 1); if (!path) { printf("Error\n"); return 0; } strcpy(path, home); printf("path = %s\n", path); // if you want to have argv[1] concatenated with path if (argc >= 2) { path = malloc(strlen(home) + strlen(argv[1]) + 1); strcpy(path, argv[1]); strcat(path, home); printf("%s\n", path); } // if you want an array of strings, each containing path, argv[1]... char **array; int i; array = malloc(argc * sizeof(char*)); array[0] = malloc(strlen(home) + 1); strcpy(array[0], home); printf("array[0] = %s\n", array[0]); for (i = 1; i < argc; i++) { array[i] = malloc(strlen(argv[i]) + 1); strcpy(array[i], argv[i]); printf("array[%d] = %s\n", i, array[i]); } // now array[i] will hold path and all the argv strings return 0; } Just as above, your path[51] is a string while path[1] is only a character, so you can't use strcpy for that. Excellent point. I've basically fixed my problem by reading up on pointers again (haven't done any C for a little while, so forgot some stuff), and doing: Code: path[1] = *home; the code doesn't moan at me when I compile it, and it runs okay (for paths which aren't close to 51 at least), but after reading what you read, I just wrote a quick program and found out that getenv("HOME") is 10 characters long, not 1 like I seem to have assumed, so I'll modify my code to fix that.
Yes, getenv will return the path to your home dir, for example /home/user, but path[1] = *home will still assign the first character of home to path[1] (which would be '/').
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Assignment makes pointer from integer without a cast
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I keep getting this error message: "warning: assignment makes pointer from integer without a cast" for line 17 (which I've made red). I've gotten this message a few times before, but I can't remember how I fixed it, and I can't figure out how to fix this one. Any suggestions? Also, it's a code that will take a password entered by the user and then run several for loops until it matches the password. It prints what it's figured out each time it guesses a new letter. Code: #include <stdio.h> #include <string.h> int main(void) { int i, j; char password[25]; char cracked[25]; char *p; char guess = '!'; printf("Enter a password of 25 characters or less: \n"); scanf("%s", password); printf("Password is being cracked..."); for (i = 0, p = password[i]; i < 25; i++, p++) { for(j = 0; j < 90; j++) { if (*p == guess) { strcpy(p, cracked); printf("\t %s \n"); break; } guess++; } //end <search> for loop } //end original for loop return 0; }
Last edited by deciel; 12-13-2011 at 01:57 AM .
Code: for (i = 0, p = password[i]; i < 25; i++, p++) password[i] is the value at index i of password . You want the address of said value, so you want p = &password[i] (or, equivalently, p = password + i ).
Oh! Thank you, it worked!
Originally Posted by deciel I keep getting this error message: "warning: for (i = 0, p = password[i]; i < 25; i++, p++) [/CODE] Note: password[i] == password[0] == *password since this is the assignment portion of for loop and i is set to zero (0).
If you want to set a pointer to the beginning of an array, just use Code: p = password An array name is essentially a pointer to the start of the array memory. Note, for a null terminated string, you could just test for Code: *p //or more explicitly *p == '\0' Also, a 25-element char array doesn't have room for a 25 character string AND a null terminator. And, ask yourself, what's going on when I enter, say a 10 character password, and i > 10.
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My C program returns error? Warning: initialization makes integer from pointer without a cast [-wint-conversion]|? help!
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This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL) 
C语言assignment makes pointer from integer without a cast 这个警告的意思是将一个int整数值直接赋值给了一个指针变量。（ 重点是类型不一致 ） 消除警告的方法就是明确类型转换是否是正确的，如果确实要把整数变量赋予指针变量，那么请使用强制类型转换。否则，请用相同的数据类型，这样编译器就不会显示警告。 比如： int *p = 10; //这就会产生这个警告 //因为 p 是指针变量，存放的是地址。而10是一个整数常量 改成： int *p = (int *)10 //强制转换成同一类型就可以消除警告 //强制类型转换，10强制转换成了一个地址  请填写红包祝福语或标题  你的鼓励将是我创作的最大动力  您的余额不足，请更换扫码支付或 充值  1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。 2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。  
compiler warning: pointer from integer without a castWould appreciate some understanding of a compiler warning. Developing for nRF52832 using SDK 15.3.0 plus S112. This code compiles without warning: ret_code_t err_code; err_code = nrfx_ppi_channel_assign(chan_0, nrfx_gpiote_in_event_addr_get(COMPARATOR_PIN),nrfx_timer_task_address_get(&m_timer1,NRF_TIMER_TASK_START)); APP_ERROR_CHECK(err_code); Then I switch from nrfx to sd calls: ret_code_t err_code; err_code = sd_ppi_channel_assign(chan_0, nrfx_gpiote_in_event_addr_get(COMPARATOR_PIN),nrfx_timer_task_address_get(&m_timer1,NRF_TIMER_TASK_START)); APP_ERROR_CHECK(err_code); and the following warnings: passing argument 2 of 'sd_ppi_channel_assign' makes pointer from integer without a cast [-Wint-conversion] passing argument 3 of 'sd_ppi_channel_assign' makes pointer from integer without a cast [-Wint-conversion] All seems to work, but would appreciate understanding the warnings. Many thanks,
Top Replies Take a look at the API documentation for explanation: nrfx_ppi_channel_assign sd_ppi_channel_assign The second one takes a pointer to a variable, while the first one takes a value. I would have call nrfx_gpiote_in_event_addr_get and nrfx_timer_task_address_get first and saved the results in their own variables that you can then pass via a pointer (&) to the sd_ppi_channel_assign. It's a bit surprising that you were able to compile with the wrong argument type. Pointers are the numbers containing the memory address of a variable instead of being the variable themselves. Therefore it could technically work, but would be very likely to behave incorrectly or even crash due to wrong values. That's what the compiler is warning you about. Best regards,
Thank you Marjeris. Makes sense. I incorrectly assumed the function parameters were the same. However, it's odd that it worked. Compiler warnings went away, but PPI stopped working when I changed the code to: ret_code_t err_code; uint32_t event_addr = nrfx_gpiote_in_event_addr_get(COMPARATOR_PIN); uint32_t task_addr = nrfx_timer_task_address_get(&m_timer1,NRF_TIMER_TASK_START); err_code = sd_ppi_channel_assign(chan_0, &event_addr, &task_addr); APP_ERROR_CHECK(err_code); I need to debug and see if sd_ppi_channel_assign returns an error. Will do so, but thought to check if the above code looks correct to you. Also, when I search devzone, all posts with sd_ppi_channel_assign access peripherals (TIMER, GPIOTE, etc.) directly, except for this post , which casts nrf_drv_gpiote_in_event_addr_get and nrf_drv_timer_task_address_get to (const volatile void *). This resolves the compiler warnings, but would like to confirm this is correct code.  The correct approach is to cast the type (const volatile void *) when providing the variables (or events/task addresses). The sd_ppi_channel_assign () function is taking pointers, but the event/task addresses can be seen as pointers itself. It is therefore not correct to pass a pointer to the variables, as this will lead to PPI endpoints being set to addresses in RAM, which are not supported. ret_code_t err_code; uint32_t event_addr = nrfx_gpiote_in_event_addr_get(COMPARATOR_PIN); uint32_t task_addr = nrfx_timer_task_address_get(&m_timer1,NRF_TIMER_TASK_START); err_code = sd_ppi_channel_assign(chan_0, (const volatile void *)event_addr, (const volatile void *)task_addr); APP_ERROR_CHECK(err_code); Best regards, Jørgen Thank you Jørgen. I think I understand. Am I correct to say that this: is equivalent to: ret_code_t err_code; err_code = sd_ppi_channel_assign(chan_0, (const volatile void *)nrfx_gpiote_in_event_addr_get(COMPARATOR_PIN), (const volatile void *)nrfx_timer_task_address_get(&m_timer1,NRF_TIMER_TASK_START)); Any advantages of one compared to the other?
Collectives™ on Stack OverflowFind centralized, trusted content and collaborate around the technologies you use most. Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Get early access and see previews of new features. assignment to 'int' from 'void *' makes integer from pointer without a castI am trying to make a linked list stack, and I followed a tutorial online, however I get this warning, and the author does not. During debugging I get a segmentation fault, if I don't use malloc in the initialize function. MINGW-W64 compiler. 
2 Answers 2The warning is due to the fact the the NULL macro is defined (by most modern compilers) as ((void *)0) , as it is intended to be used only for pointers. Assigning this value to the data member of your structure causes the warning. To remove this warning, use s->data=0; in place of s->data=NULL; . The author of the tutorial is either using an older compiler or has disabled this particular warning. Also, the pointer ( s ) passed to your Initialize function will be a copy of whatever variable you use as an argument when you call that function and, as such, its value is not updated in the calling code. You haven't specified how you intend to use that function, but here is a (perhaps) better implementation: And then, when you call that function, you can assign its returned value to your 'global' top pointer: 
In C the macro NULL is defined like that is the type of the expression is a null pointer of the type void * . Thus in this statement a pointer is assigned to an object of the type int and the compiler issues a message saying that you are doing something wrong. The function Initialize as it is written does not make a sense. Apart of all for example this statement does not change the original pointer top used as the function argument. In fact the function is redundant. Nevertheless if to write such a function then it can look either like and called like Or for a new created node it can be declared and defined like  Your AnswerReminder: Answers generated by artificial intelligence tools are not allowed on Stack Overflow. Learn more Sign up or log inPost as a guest. Required, but never shown By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy . Not the answer you're looking for? Browse other questions tagged c pointers memory struct or ask your own question .
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line[100]で 「1」が格納されていたら「a」 「2」が格納されていたら「b」 「3」が格納されていたら「c」 とout[100]に代入する関数を作りたいのですが コンパイルすると関数の部分で warning: assignment makes integer from pointer without a cast という警告がでます。 ポインターは使っていないのですが、ポインターに関する警告が出ているようで困っています。 どこが悪いのかまったくわからなくて作業が完全に止まってしまいました。 解決法をおしえてください。お願いします。 /*宣言*/ int=i; /*main関数内のfor文で使用*/ char line[100], out[100]; void change(int); /*関数*/ void change(int i) { if(line[i]=='1'){ out[10]="a\0" }if(line[i]=='2'){ out[10]="b\0"; }if(line[i]=='3'){ out[10]="c\0" } } この質問への回答は締め切られました。 A 同じカテゴリの新着質問 A 回答 (2件)  No.1 ベストアンサー
＞ out[10]="a\0" 回答ありがとうございます。 "a\0"という書き方がまずかったのですか。 以後気をつけます。 
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However, this returns a char. So your assignment. cString1 = strToLower(cString1); has different types on each side of the assignment operator .. you're actually assigning a 'char' (sort of integer) to an array, which resolves to a simple pointer. Due to C++'s implicit conversion rules this works, but the result is rubbish and further access to ...
I need to remove all the commas from my user input. The code is working but it's giving the warning "assignment to 'char' from 'char *' makes integer from pointer without a cast". I need to get rid of this warning.
1. Earlier, I asked a similar question, but I've since changed my code. Now the compiler gives me a different warning. This is an example of what my code looks like now: void *a = NULL; void *b = //something; a = *(int *)((char *)b + 4); When I try to compile, I get "warning: assignment makes pointer from integer without a cast."
Converting Integers to Pointers {#converting-integers-to-pointers} To convert an integer to a pointer, follow these steps: Include the <stdint.h> header (C) or the <cstdint> header (C++) in your program. Cast your integer to the required pointer type using a double cast. int a = 42; uintptr_t int_ptr = (uintptr_t)&a;
Solving Assignment Makes Integer from Pointer Without a Cast in C: Tips and Solutions
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Re: C - assigment makes integer from pointer without a cast warning. path [1] is the second element of a char array, so it's a char. home is a char *, i.e. a pointer to a char. You get the warning because you try to assign a char* to a char. Note also the potential to overflow your buffer if the content of the argv [1] argument is very long.
Assignment makes pointer from integer without a cast Getting started with C or C++ | C Tutorial | C++ Tutorial | C and C++ FAQ | Get a compiler | Fixes for common problems Thread: Assignment makes pointer from integer without a cast
initialization makes pointer from integer without a cast You are initializing your reversed_arr variable, which is a pointer to a char, with the result of the reverse function, which returns a char (which is basically an integer), which is why you are getting your warning. You could try: char *reversed_arr = (char *) reverse(arr);
How To Stop a Pointer Creation From an Integer Without a Cast. - Use Equal Data Types During Assignment. - Ensure the Pointer and Integer Have the Same Sizes. - Pass a "Format String" to the "Printf ()" Function. - Use a Function That Returns a Pointer to "Struct _IO_file". - Copy the String to Character Array or Character ...
Strings literals are pointers in C. When you a sentence like "Hello world!", that variable is of the type const char*, meaning it is a pointer to a bunch of chars that cannot be modified. When you use double-quotes, you are making a const char*, even if there is only one character. You should use single-quotes if you want to have a char.
To add onto this, if you're wanting to compare a single character (here '-' and '+'), you'll want to use single quotes rather than double quotes. Single quotes indicate characters (char), whilst double quotes indicate null-terminated strings (char *).
warning: initialization makes integer from pointer without a cast [-Wint-conversion]| What I have tried: I have tried stackoverflow,I have also looked all over google but couldnt find the awnser.I also tried quora
In this case a[4] is the 5th integer in the array a, ap is a pointer to integer, so you are assigning an integer to a pointer and that's the warning. So ap now holds 45 and when you try to de-reference it (by doing *ap) you are trying to access a memory at address 45, which is an invalid address, so your program crashes.. You should do ap = &(a[4]); or ap = a + 4;
text should be declared as: char *text = NULL; You need a pointer to char. This pointer can be set to point to any literal string (as you do in your case statements). char text; // this is just a single character (letter), not a string. 2. Objective_Ad_4587 • 3 yr. ago. i got it thank you very much.
近日遇见一个bug，最后调查是程序的warning引起的： 编译的时候报警告： assignment makes pointer from integer without a cast 出现这个警告的原因是在使用函数之前没有对函数进行声明，未经声明的函数原型一律默认为返回int值。就相当于你调用了返回值为int的函数，并将其赋给了char*变量，所有会出现警告。
passing argument 3 of 'sd_ppi_channel_assign' makes pointer from integer without a cast [-Wint-conversion] All seems to work, but would appreciate understanding the warnings. Many thanks, Tim. Hi, From what I can tell, the two are equivalent (except for the second one missing the call to APP_ERROR_CHECK ()).
OK - in that case, since you want to immediately change the value of sad, it can be a local variable to the function, and doesn't need to be an argument to the function: . void racun(int D, int M, int G, int *dUg) { int i, *sad; If you want sad to point to an array, such that accessing sad (sad[...]) will be equivalent to accessing that array (e.g. obicna[...]), you can simply assign the array ...
Either there is something wrong in function Initialize or the function argument stack *s is useless and should be replaced with a local variable. The value s passed to the function is immediately overwritten with the result of malloc.The caller of Initialize will not get the modified value as the pointer s is passed by value. 2nd problem: You should check the return value of malloc.
warning: assignment makes integer from pointer without a cast という警告がでます。 ポインターは使っていないのですが、ポインターに関する警告が出ているようで困っています。 どこが悪いのかまったくわからなくて作業が完全に止まってしまいました。
How to fix "warning: assignment to 'int (*)(int, int, int, void *)' from 'int' makes pointer from integer without a cast [-Wint-conversion]" I am currently reading a book on Ethical Hacking and I have hit a road block with this warning. The book uses C for this example in Sys call hooking, and I seem to keep getting this warning message ...