As others already noted, in one case you are attempting to return cString (which is a char * value in this context — a pointer) from a function that is declared to return a char (which is an integer). In another case you do the reverse: you are assigning a char return value to a char * pointer. This is what triggers the warnings. You certainly need to declare your return values as char *, not as char.
Note BTW that these assignments are in fact constraint violations from the language point of view (i.e. they are «errors»), since it is illegal to mix pointers and integers in C like that (aside from integral constant zero). Your compiler is simply too forgiving in this regard and reports these violations as mere «warnings».
What I also wanted to note is that in several answers you might notice the relatively strange suggestion to return void from your functions, since you are modifying the string in-place. While it will certainly work (since you indeed are modifying the string in-place), there’s nothing really wrong with returning the same value from the function. In fact, it is a rather standard practice in C language where applicable (take a look at the standard functions like strcpy and others), since it enables «chaining» of function calls if you choose to use it, and costs virtually nothing if you don’t use «chaining».
That said, the assignments in your implementation of compareString look complete superfluous to me (even though they won’t break anything). I’d either get rid of them
int compareString(char cString1[], char cString2[]) {
// To lowercase
strToLower(cString1);
strToLower(cString2);
// Do regular strcmp
return strcmp(cString1, cString2);
}
or use «chaining» and do
int compareString(char cString1[], char cString2[]) {
return strcmp(strToLower(cString1), strToLower(cString2));
}
(this is when your char * return would come handy). Just keep in mind that such «chained» function calls are sometimes difficult to debug with a step-by-step debugger.
As an additional, unrealted note, I’d say that implementing a string comparison function in such a destructive fashion (it modifies the input strings) might not be the best idea. A non-destructive function would be of a much greater value in my opinion. Instead of performing as explicit conversion of the input strings to a lower case, it is usually a better idea to implement a custom char-by-char case-insensitive string comparison function and use it instead of calling the standard strcmp.
Makes pointer from integer without a cast is among the errors that you’ll see from your C compiler when you perform an invalid operation with a pointer. This article is your ultimate collection of code samples that will lead to these types of errors, and we’ll teach you how you can fix them.
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.
Contents
- Why Your Code Made a Pointer From an Integer Without a Cast?
- – 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
- 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 Pointer
- – Assign Pointers of Compatible Types
- Conclusion
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
– You Assigned an Integer to a Pointer
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<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();
}
– You Want To Convert an Integer to a Pointer
Converting an integer to a pointer leads to the “cast to pointer from integer of different size” error.
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.
int main() {
int theta = 10;
int *ptr_theta = (int *)theta;
}
– You Passed a Variable Name to the “Printf()” Function
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.
int main() {
int num_val = 42;
printf(num_val); //Error!
return 0;
}
– You Used “Struct _IO_file” in a Wrong Way
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:
int main(void) {
int alpha = 10;
FILE *beta = alpha;
return 0;
}
– You Copied a String to an Invalid Location
The “strcpy()” expects a valid location when you’re copying strings. If you fail to do this, your C compiler will throw the “strcpy makes pointer from integer without a cast error.
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<string.h>
int main() {
int destination;
char source[] = “Hello, World!”;
strcpy(destination, source);
return 0;
}
– You’re Setting a Pointer to a Different Type
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.
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”.
The following methods also work:
- Copy the string to a character array or character pointer
- Assign pointers of compatible types
– Use Equal Data Types During Assignment
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<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]);
}
}
– Ensure the Pointer and Integer Have the Same Sizes
To convert an integer to a pointer using a typecast, ensure they have the same sizes. We did not do this in our second example in this article, and it led to an error.
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<stdint.h>
int main() {
int theta = 10;
int *ptr_theta = (int *)(intptr_t)theta;
printf(“%p”, ptr_theta);
}
– Pass a “Format String” to the “Printf()” Function
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”.
int main() {
int num_val = 42;
printf(“%d”, num_val);
return 0;
}
– Use a Function That Returns a Pointer to “Struct _IO_file”
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”.
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;
}
– Copy the String to Character Array or Character Pointer
To copy a string using the “strcpy()” function, ensure that0 “destination” is a character array or a character pointer.
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<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<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;
}
– Assign Pointers of Compatible Types
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.
int main() {
int charlie = 5;
int *pacifier = &charlie;
int *qwerty = pacifier;
}
Conclusion
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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In C programming, one of the common warnings you might encounter is the «warning: assignment makes pointer from integer without a cast» warning. This warning typically occurs when you try to assign an integer value to a pointer variable without explicitly casting it. In this guide, we will take a look at the causes of this warning and walk you through the steps to fix it.
Table of Contents
- Understanding the Warning
- Step-by-Step Guide to Fix the Warning
- FAQs
- Related Links
Understanding the Warning
Let’s first understand the root cause of this warning. The C programming language is strongly typed, which means that the type of a variable must be explicitly defined. When you assign a value to a pointer, the type of the value being assigned must match the type of the pointer. If the types don’t match, the compiler generates a warning. This is what happens when you see the «assignment makes pointer from integer without a cast» warning.
Consider the following example:
int main()
{
int *ptr;
ptr = 10;
return 0;
}
In this example, the int *ptr declares a pointer to an integer. However, when we try to assign the value 10 to the pointer, the compiler generates the warning because it is attempting to assign an integer value to a pointer.
Step-by-Step Guide to Fix the Warning
Step 1: Identify the Problematic Line of Code
The first step to fixing this warning is to identify the line of code that is causing the problem. The compiler will usually provide the line number where the warning occurs. For example:
warning: assignment makes pointer from integer without a cast [-Wint-conversion]
In our example above, the problematic line of code is ptr = 10;.
Step 2: Check If an Explicit Cast Is Required
In some cases, you might need to convert the integer value to a pointer using an explicit cast. To do this, you can use the following syntax:
pointer_variable = (pointer_type) integer_value;
For example, if you want to assign the address of the integer value 10 to the pointer ptr, you can use an explicit cast as shown below:
int main()
{
int *ptr;
ptr = (int *) 10;
return 0;
}
However, it is essential to ensure that the explicit cast is correct and meaningful in the context of your program.
Step 3: Correct the Assignment
In most cases, the warning occurs because you are trying to assign an integer value to a pointer, which is not the correct way to initialize a pointer. Instead, you should assign the address of a variable to the pointer. For example:
int main()
{
int *ptr;
int variable = 10;
ptr = &variable;
return 0;
}
In this example, we have declared an integer variable variable and assigned the value 10 to it. We then assign the address of variable to the pointer ptr using the & operator.
By following these steps, you should be able to fix the «assignment makes pointer from integer without a cast» warning in your C program.
FAQs
1. What is the role of pointers in C programming?
Pointers are variables that store the address of another variable. They are used for various purposes, such as accessing memory locations, passing function arguments by reference, and implementing data structures like linked lists and trees.
2. What is the difference between a pointer and a reference?
A pointer is a variable that stores the address of another variable, while a reference is an alias for an existing variable. References cannot be reassigned, whereas pointers can be.
3. What is casting in C programming?
Casting is the process of converting a variable from one data type to another. It can be done explicitly using a cast operator or implicitly by the compiler.
4. How do I declare a pointer to a function in C?
To declare a pointer to a function, use the following syntax:
return_type (*pointer_name)(parameter_type);
For example, to declare a pointer to a function that takes an integer as a parameter and returns an integer, you would write:
int (*function_ptr)(int);
5. What is the purpose of the sizeof operator in C?
The sizeof operator is used to determine the size (in bytes) of a variable or data type. It can be helpful when working with memory allocation and pointer arithmetic.
- Pointers in C – A Hands-on Approach
- Casting in C Programming
- C Function Pointers
Дословно это можно перевести как «присваивание делает/формирует указатель из целого [значения] без [использования] явного приведения типа».
Термин «cast» в С означает явное приведение типа. Именно явное.
Сочетание «makes […] without a cast» ссылается на то, что ваш оператор присваивания пытается выполнить именно неявное преобразование из указателя к целому значению.
Все сообщение означает, что в своем коде использован оператор присваивания (assignment), правая часть которого является указателем (pointer), а левая — целым числом (integer).
Такой код является некорректным в С, ибо язык С в общем случае запрещает неявные преобразования между целыми числами и указателями. Чтобы преобразовать целое значение к указателю в С требуется явное приведение типа, т.е. cast.
Как правило встречаются такие способы передачи данных из функции.
Во-первых, можно в функции создать статический массив в функции, а потом возращать указатель на него. Как то так:
Код: Выделить всё
#define MAXLINE 100
char* get_name_addition(void)
{
static char name[MAXLINE];
fgets(name,MAXLINE,stdin);
... // Какие-то дополнительные действия
return *name;
}Типичная практика для Unix-а (насколько могу судить), но, как нетрудно заметить, это функция не совместима с многопоточным программированием. Эту функцию можно переделать для многопоточных приложений, используя собственные данные потоков (Thread-Specific Data), но лучше сразу делать иначе.
Пожалуй самая распространенная практика — это передать в функцию буфер для записи — этот буфер не будет привязан к стеку выполняемой функции.
Код: Выделить всё
#define MAXLINE 100
int get_name_addition(char *name, size_t n)
{
fgets(name, n, stdin);
...
return 0;
}Не всегда можно заранее угадать, сколько нужно выделить места для буфера — это подчас известно только в вызываемой функции. К примеру в Win эта задача часто решается повторным вызовом одной функции: первый раз задаем нулевой буфер и узнаем необходимый размер; второй раз вызывает функцию с буфером достаточного размера и получаем в него данные. Здесь, в никсах, мне такого не попадалось.
Но встречается более простой способ. В вызываемой функции malloc-ом создаем буфер. Но для его очистки создаем еще одну дополнительную функцию.
Код: Выделить всё
#define MAXLINE 100
char* get_name_addition(void)
{
char* name;
name = (char*)malloc(MAXLINE);
fgets(name, MAXLINE, stdin);
...
return name;
}
void get_name_free(char* name)
{
free(name);
}Соответственно, использовать это надо так:
Код: Выделить всё
int main(void)
{
char *name;
name = get_name_addition();
// Что-то делаем
...
get_name_free(name);
return 0;
}Конечно, в данном случае можно было бы в main-е просто вызвать free. Но, во первых, возращаемые данные не обязательно будут линейным куском памяти, а могут оказаться каким нибудь связным списком. Во-вторых, если функция та находится в разделяемой библиотеке, я бы не стал исключать случай, что там могла быть слинкована другая реализация malloc-а.