Question

我对以下程序有疑问：它打印：

dst-＆gt; val in f1 = 6

dst.val in main = -528993792

我想修复此程序，以便打印

<= dst.val in main = 6

我该怎么做？

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

typedef struct my_struct myStruct;
struct my_struct
{
    int val;
};

myStruct *f2(void)
{
    myStruct *dst = malloc(sizeof(myStruct));
    dst->val = 6;
    return dst;
}



void f1(myStruct *dst)
{
    dst = f2();
    printf("**dst->val in f1=%d\n", dst->val);
}


int main()
{
    myStruct dst;
    f1(&dst);
    printf("**dst.val in main=%d\n", dst.val);
}

Answer 1

按值返回结构;不要使用指针和动态分配：

myStruct f2(void)
{
  myStruct dst;
  dst.val = 6;
  return dst;
}

然后f1将更常规地使用pass-by-pointer概念：

void f1(myStruct *dst)
{
    *dst = f2();
    printf("**dst->val in f1=%d\n", dst->val);
}

（实际上，我不确定它是否被称为“传递指针”或“传递参考”;可能后者是正确的）

Answer 2

void f1(myStruct *dst){
    myStruct *p = f2();
    printf("**dst->val in f1=%d\n", p->val);
                      //dst is the address that was copied on the stack.
    dst->val = p->val;//There is no meaning If you do not change the contents.
    free(p);
}

我该怎么做来修复这个程序？

2 个答案: