为什么这会导致 C 中的分段错误

Question

I try to put my pointer to the parameter passed it, but it leads to a segmentation fault

#include <stdio.h>

void putPointer ( int *point, int num ) {
    point = &num;
}

int main(void) {

    int a = 42;

    int *p;
    // p = &a
    putPointer( p, a );

    printf( "%d\n", *p );

    return 0;
}

Answer 1

Arguments in C are copies of what are passed, so p in the main() function remains uninitialized after putPointer(p,a); . Using values of uninitialized non-static local variables, which are indeterminate, invokes undefined behavior .

Also note that you shouldn't take out pointers to non-static local variables (including function arguments) on returning from functions and must not use these pointers after returning from functions because non-static local variables are invalidated on returning.

Answer 2

In putPointer() you try to assign point but point is a copy of p , thus p is left unchanged. Since p is still not initialised, it could access any address and *p causes a seg-fault.

You could actually change p by expecting int **point_ptr as parameter, passing &p and performing *point_ptr=# but you would encounter another problem. Actually, num will disappear as soon as the function returns, so keeping its address in p is incorrect: num does not exist anymore when using *p .

Answer 3

&num

will be the address of num on the stack . You are trying dereference the stack in printf("%d\n", *p); .