如何将函数内部的int指针传递给另一个函数并在那里操作in？

Question

#include <stdio.h>

void testFunction(int *a){
    printf("In a should be 5: %d", *a);
    *a = 0;
    printf("In a should be 0: %d", *a);
    test2Function(&a);
}

void test2Function(int *a){
    printf("In F2 a should be 0: %d", *a);
    *a = 2;
    printf("In F2 a should be 2: %d", *a);
}

int main(){

    int a = 5;
    testFunction(&a);
    printf("In main() a should be 2: %d", a);


    return 0;
}

如何将指针从给定函数发送到函数内的函数到另一个函数？所以上面代码中的语句都是真的

目前的输出是:

In a should be 5: 5
In a should be 0: 0
In F2 a should be 0: 6422316
In F2 a should be 2: 2
In main() a should be 2: 0

Answer 1

代替:

test2Function(&a);

做:

test2Function(a);

因为您只想复制已有的指针,而不是获取指针本身的地址.