C 指针问题 - 局部变量的地址可能会转义函数

Question

Here is my code:

int* return_to_main()
{
    int george = 32;
    int* pointer;
    pointer = &george;
    printf("george address: %p\n", (void *)&george);
    printf("george content: %d\n", *pointer);
    return pointer;
}

int main()
{
    int* test = return_to_main();
    printf("test address: %p\n", (void *)&test);
    printf("test content: %d\n", *test);
}

and it returns:

george address: 0061FEE8
george content: 32
test address: 0061FF1C
test content: 6422476

CLion says that "Address of local variable may escape the function". How do I fix this? And why is variable george and test's address and contents different? Is it normal that variable george and variable test's address are different???

Answer 1

"Address of local variable may escape the function".

The reason for that is you are returning the address of local variable, you should not do that instead return by value or use dynamic memory allocation/ global variable etc.

one way to resolve your warning is below

#include <stdio.h>
#include <stdlib.h>

int* return_to_main()
{
    int george = 32;
    int* pointer;
    pointer = malloc(sizeof(int));
    if ( !pointer ) {
        printf("malloc failed\n");
        return NULL;
     }
    *pointer = george;
    printf("george address: %p\n", (void *)&george);
    printf("pointer : %p\n", pointer);
    printf("george/pointer content: %d\n", *pointer);
    return pointer;
}

int main()
{
    int* test = return_to_main();

    if(test)
    {
        printf("test address: %p\n", (void *)test);
        printf("test content: %d\n", *test);
        free(test);
        test = NULL;
    }
    return 0;
}