Question

我不太清楚这段代码是如何运作的：

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我的理解是char在内存中有1个字节（大小），我们使用指针交换int值的每个字节（4个字节）。
但最后，如何将char指针取消引用到int值？

Answer 1

让我们一步一步地尝试用代码注释

来解决这个问题

#include <stdio.h>

//gswap() takes two pointers, prta and ptrb, and the size of the data they point to
void gswap(void* ptra, void* ptrb, int size)
{
    // temp will be our temporary variable for exchanging the values
    char temp;
    // We reinterpret the pointers as char* (byte) pointers
    char *pa = (char*)ptra;
    char *pb = (char*)ptrb;
    // We loop over each byte of the type/structure ptra/b point too, i.e. we loop over size
    for (int i = 0 ; i < size ; i++) {
        temp = pa[i]; //store a in temp
        pa[i] = pb[i]; // replace a with b
        pb[i] = temp; // replace b with temp = old(a)
    }
}

int main()
{
    // Two integers
    int a=1, b=5;
    // Swap them
    gswap(&a, &b, sizeof(int));
    // See they've been swapped!
    printf("%d , %d", a, b);
}

所以，基本上，它的工作原理是遍历任何给定的数据类型，重新解释为字节，并交换字节。

在C中使用指向char的指针的通用交换函数

1 个答案: