assignment to int from int makes pointer from integer without a cast wint conversion

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

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How to Fix Initialization Makes Pointer from Integer Without a Cast Error

If you are a developer, you might have come across the error message "initialization makes pointer from integer without a cast" during your programming career. This error message typically occurs when you try to assign an integer value to a pointer variable without explicitly casting the integer to a pointer. In this guide, we will discuss the causes of this error and provide you with a step-by-step solution to fix it.

Understanding the 'Initialization Makes Pointer from Integer Without a Cast' Error

Before we dive into the solution, let's first understand what this error message means. In C and C++, pointers are variables that hold memory addresses. They are used to access memory locations and manipulate data stored in those locations. When you assign an integer value to a pointer variable, the compiler treats it as an attempt to create a pointer from an integer, which is not allowed in C and C++. This is because pointers and integers have different data types and sizes.

The error message "initialization makes pointer from integer without a cast" is a compiler error that occurs when you try to assign an integer value to a pointer variable without explicitly casting the integer to a pointer type. This error message is typically accompanied by a line number and a file name that indicates where the error occurred.

Causes of the 'Initialization Makes Pointer from Integer Without a Cast' Error

The 'Initialization Makes Pointer from Integer Without a Cast' error can occur due to various reasons. Some of the common causes are:

Assigning an integer value to a pointer variable without casting it to a pointer type.
Using an incorrect data type for the pointer variable.
Declaring a pointer variable without initializing it.
Passing an integer value to a function that expects a pointer.

Solution to the 'Initialization Makes Pointer from Integer Without a Cast' Error

To fix the 'Initialization Makes Pointer from Integer Without a Cast' error, you need to cast the integer value to a pointer type before assigning it to a pointer variable. Here are the steps to fix this error:

Identify the line of code that is causing the error. The error message typically indicates the line number and the file name where the error occurred.
Check if the variable on the left-hand side of the assignment operator is a pointer variable. If it is not a pointer variable, you need to declare it as a pointer variable.
Cast the integer value to a pointer type using the appropriate casting operator. The casting operator in C and C++ is the (type) operator. For example, if you want to cast an integer value to a pointer to an integer, you can use the following syntax:
Compile and run your code to verify that the error has been fixed.

Frequently Asked Questions (FAQ)

Q1. what is a pointer in c and c++.

A pointer is a variable that holds the memory address of another variable. Pointers are used to access memory locations and manipulate data stored in those locations.

Q2. What is the data type of a pointer in C and C++?

The data type of a pointer depends on the data type of the variable it points to. For example, a pointer that points to an integer variable has the data type 'int *'.

Q3. What is a casting operator in C and C++?

A casting operator is an operator that is used to convert one data type to another. In C and C++, the casting operator is the (type) operator.

Q4. How do I declare a pointer variable in C and C++?

To declare a pointer variable in C and C++, you need to use the '*' operator. For example, to declare a pointer to an integer variable, you can use the following syntax:

Q5. Can I assign a pointer to an integer variable?

No, you cannot assign a pointer to an integer variable without casting it to an integer type. This is because pointers and integers have different data types and sizes.

Makes Pointer From Integer Without a Cast: Fix It Now!

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Makes Pointer From Integer Without a Cast

Our research team has ensured that this will be your best resource on these error messages, and you won’t have to read another article about it again. With that said, launch your code that’s showing the error messages, and let’s fix it for you.

JUMP TO TOPIC

– You Assigned an Integer to a Pointer

– you want to convert an integer to a pointer, – you passed a variable name to the “printf()” function, – you used “struct _io_file” in a wrong way, – you copied a string to an invalid location, – you’re setting a pointer to a different type, – 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 pointer, – assign pointers of compatible types, why your code made a pointer from an integer without a cast.

Your code made a pointer from an integer without a cast because you assigned an integer to a pointer or you want to convert an integer to a pointer. Other causes include the passing of a variable name to the “printf()” function and using “struct _io_file in the wrong way.

Finally, the following are also possible causes:

You copied a string to an invalid location
You’re setting a pointer to a different type

If you assign an integer to a pointer, it will lead to the “ assignment makes pointer from integer without a cast c programming ” error. For example, in the following, the “num_array” variable is an unsigned integer type while “tmp_array” is a pointer.

Later, an error will occur when the “for” loop tries to copy the values of the “num_array” to the “tmp_array” pointer using a variable assignment.

#include<stdio.h>

#include<stdint.h>

uint8_t num_array[8] = {0, 1, 2, 3, 4, 5, 6, 7};

void copy_arr_values() {

int i;

uint8_t *tmp_array[8];

for(i=0; i<8; i++){

tmp_array[i] = num_array[(i+3)%8];

}

int main() {

copy_arr_values();

}

Makes Pointer From Integer Without a Cast Causes

For example, in the following, the “theta” variable is an integer while “ptr_theta” is a pointer. Both have different sizes, and you can’t convert the integer to a pointer.

#include<stdio.h>

int main() {

int theta = 10;

int *ptr_theta = (int *)theta;

}

If you pass a variable name to the “printf()” function, that’s when the compiler will throw the “ passing argument 1 of ‘printf’ makes pointer from integer without a cast ” error.

For example, in the following, “num_val” is an integer variable, and the code is trying to print it using the “printf()” function.

#include <stdio.h>

int main() {

int num_val = 42;

printf(num_val); //Error!

return 0;

}

When you assign an integer to a pointer of “struct _io_file”, that’s when you’ll get the “ assignment to file aka struct _io_file from int makes pointer from integer without a cast ” error. For example, in the following code, “FILE” is a “typedef” for “struct _io_file”. This makes it a pointer to “struct _io_file”.

Later, the code assigned it to the integer “alpha,” and this will lead to an error stated below:

#include <stdio.h>

int main(void) {

int alpha = 10;

FILE *beta = alpha;

return 0;

}

Makes Pointer From Integer Without a Cast Reasons

Now, in the following, the code is trying to copy a string (“source”) into an integer (“destination”), and this leads to an error because it’s not a valid operation:

#include<stdio.h>

#include<string.h>

int main() {

int destination;

char source[] = “Hello, World!”;

strcpy(destination, source);

return 0;

}

When your code sets a pointer to a different type, your compiler will show the “ incompatible pointer type ” error. In the following code, “pacifier” is an integer pointer, while “qwerty” is a character pointer.

Both are incompatible, and your C compiler will not allow this or anything similar in your code.

#include<stdio.h>

int main() {

int charlie = 5;

int *pacifier = &charlie;

char *qwerty = pacifier;

}

How To Stop a Pointer Creation From an Integer Without a Cast

You can stop a pointer creation from an integer without a cast if you use equal data types during the assignment or ensure the integer and pointer have the same sizes. What’s more, you can pass a “format string” to “printf()” and use a function that returns a pointer to “struct _io_file”.

Copy the string to a character array or character pointer
Assign pointers of compatible types

During a variable assignment, use variables of equal data types. In our first example, we assigned a pointer (uint8_t *tmp_array) to an unsigned integer (uint8_t num_array) and it led to a compile-time error.

Now, the following is the revised code, and we’ve changed “tmp_array” to a real array. This will prevent the “ gcc warning: assignment makes pointer from integer without a cast ” error during compilation.

#include<stdio.h>

#include<stdint.h>

uint8_t num_array[8] = {0, 1, 2, 3, 4, 5, 6, 7};

int main() {

int i;

// Fix: change tmp_array to an array

uint8_t tmp_array[8] = {};

for (i = 0; i < 8; i++){

tmp_array[i] = num_array[(i+3)%8];

}

int length = sizeof(tmp_array)/sizeof(tmp_array[0]);

printf(“Elements of the ‘tmp_array’: \n”);

for (int i = 0; i < length; i++) {

printf(“%d “, tmp_array[i]);

}

Makes Pointer From Integer Without a Cast Fixes

Now, the fix is to use “intptr_t” from the “stdint.h” header file because it guarantees that the pointer and integer will have the same sizes. We’ve used it in the following code, and you can compile it without an error.

#include<stdio.h>

#include<stdint.h>

int main() {

int theta = 10;

int *ptr_theta = (int *)(intptr_t)theta;

printf(“%p”, ptr_theta);

}

To fix the “pointer to integer without a cast” in the “printf()” function, pass a “format string” as its first argument. How you write the “format string” depends on the variable that you’ll print. In the following updated code, “num_val” is an integer, so the “format string” is “%d”.

#include <stdio.h>

int main() {

int num_val = 42;

printf(“%d”, num_val);

return 0;

}

When you’re using “FILE” from the “stdlib.h” header file, you can prevent any “pointer from integer” error if you use a function that returns a pointer to “struct _io_file”.

An example of such a function is “fopen()” which allows you to open a file in C programming. Now, the following is a rewrite of the code that causes the pointer error in “struct _io_file”. This time, we use “fopen()” as a pointer to “FILE”.

#include <stdio.h>

int main(void) {

FILE *f = fopen(“myfile.txt”, “r”);

if (f == NULL) {

perror(“Error opening file”);

return 1;

}

char c;

while ((c = fgetc(f)) != EOF) {

printf(“%c”, c);

}

return 0;

}

Makes Pointer From Integer Without a Cast Solutions

Now, in the following code, we’ve changed “destination” from an integer to a “character array”. This means “strcpy()” can copy a string into this array without an error.

#include<stdio.h>

#include<string.h>

int main() {

char destination[100];

char source[] = “Hello, World!”;

strcpy(destination, source);

printf(“%s”, destination);

return 0;

}

Meanwhile, the following is another version that turns the “destination” into a “character pointer”. With this, you’ll need to allocate memory using the “malloc()” function from the “stdlib.h” header file.

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

int main() {

char* destination;

char source[] = “Hello, World!”;

destination = malloc(100);

strcpy(destination, source);

printf(“%s”, destination);

free(destination);

return 0;

}

When you’re assigning pointers, ensure that they’re compatible by changing their data type . The following is the updated code for the “charlie” example , and “qwerty” is now an integer.

#include<stdio.h>

int main() {

int charlie = 5;

int *pacifier = &charlie;

int *qwerty = pacifier;

}

This article explained why your code made a pointer without a cast and how you can fix it. The following is a summary of what we talked about:

An attempt to convert an integer to a pointer will lead to the “makes pointer from integer without a cast wint conversion” error.
To prevent an “incompatible pointer type” error, don’t set a pointer to a different type.
If you’re using the “printf()” function, and you want to prevent any “pointer to integer” error, always pass a “format specifier”.

At this stage, you’ll be confident that you can work with integers and pointers in C programming without an error. Save our article, and share it with your developer communities to help them get rid of this error as well.

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Assignment makes integer from pointer without a cast [-Wint-conversion

I really don't understand why I have such error knowing that tmp and key are the same type and size.

This produces:

warning: assignment makes integer from pointer without a cast [-Wint-conversion

variable-assignment

2 Answers 2

tmp and key are the same type

NO . They are not. They both are arrays, but the datatype is different. One is a uint8_t * array, another is a uint8_t array.

1 It wasn't my downvote, but I don't see where his code uses a pointer from tmp uninitialized. He's assigning to the i th item of tmp , not dereferencing it. It's hardly undefined behavior to assign to an uninitialized pointer. – Cody Gray ♦ Commented Jan 18, 2016 at 5:16
@CodyGray You're right sir, I also got confused by the data types....Thanks for pointing that out. But then, regarding the DV, it was before I added that part, as an edit.....so, .... :( – Sourav Ghosh Commented Jan 18, 2016 at 7:51

Not clear what you want here but if you want tmp[x] to reflect the value in key[y] then

Otherwise if you want tmp to be separate, then ...

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Case ID: 254703

compiler warning: pointer from integer without a cast

Would 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,

Vote Up 0 Vote Down
Verify Answer

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?

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assignment makes integer from pointer without a cast

I'm trying to parse network packets in C as:

ip_hdr_t is a struct defines the IP header and it has a member uint8_t ihl:4 .

When compiling the code it gave warnings and errors:

I understand the first warning, but how did the second one in line 60 come? ip_hdr is uint8_t* and ip_hdr_len is uint8_t , so I expect the addition produces an uint8_t* instead of an integer ?

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COMMENTS

Assignment makes pointer from integer without cast
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 ...
C pointers and arrays: [Warning] assignment makes pointer from integer
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;
C语言assignment makes pointer from integer without a cast
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 ...
Makes Integer From Pointer Without A Cast (Resolved)
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;
Assignment makes integer from pointer without a cast in c
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 ...
Assignment makes integer from pointer without a cast and ...
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 ...
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c
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."
"assignment makes integer from pointer without a cast -wint-conversion
"assignment makes integer from pointer without a cast -wint ...
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char *word is a pointer, meaning the value contained by the word parameter is a memory address (in this example, it will be the address of the first character in a string). When you use the notation *word on line 19, you are dereferencing the word pointer, meaning it will return the value of whatever is stored at word's memory address (which is a single character).
Makes Pointer From Integer Without a Cast: Fix It Now!
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 ...
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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);
Assignment makes integer from pointer without a cast [-Wint-conversion
It wasn't my downvote, but I don't see where his code uses a pointer from tmp uninitialized. He's assigning to the i th item of tmp , not dereferencing it. It's hardly undefined behavior to assign to an uninitialized pointer.
assignment makes integer from pointer without a cast
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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 ...
assignment makes integer from pointer without a cast
When compiling the code it gave warnings and errors: network.c:51:12: warning: assignment discards 'const' qualifier from pointer target type [enabled by default] network.c:60:12: warning: assignment makes integer from pointer without a cast [enabled by default] network.c:64:26: error: invalid type argument of unary '*' (have 'int ...

C语言assignment makes pointer from integer without a cast

How to Fix Initialization Makes Pointer from Integer Without a Cast Error

Understanding the 'Initialization Makes Pointer from Integer Without a Cast' Error

Causes of the 'Initialization Makes Pointer from Integer Without a Cast' Error

Solution to the 'Initialization Makes Pointer from Integer Without a Cast' Error

Frequently Asked Questions (FAQ)

Q2. What is the data type of a pointer in C and C++?

Q3. What is a casting operator in C and C++?

Q4. How do I declare a pointer variable in C and C++?

Q5. Can I assign a pointer to an integer variable?

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