[英]Generic swap function using pointer to char in C
I don't understand so well how this code works: 我不太了解这段代码的工作方式:
#include <stdio.h>
void gswap(void* ptra, void* ptrb, int size)
{
char temp;
char *pa = (char*)ptra;
char *pb = (char*)ptrb;
for (int i = 0 ; i < size ; i++) {
temp = pa[i];
pa[i] = pb[i];
pb[i] = temp;
}
}
int main()
{
int a=1, b=5;
gswap(&a, &b, sizeof(int));
printf("%d , %d", a, b)
}
What I understand is that char has 1 byte(size) in memory and we are using pointers to swap each byte of the int value(4 bytes). 我的理解是char在内存中有1个字节(大小),我们正在使用指针交换int值的每个字节(4个字节)。
But in the end, how it is possible to dereference a char pointer to int value? 但是最后,如何将char指针取消引用到int值?
Let's try and figure this out, step by step, with code comments 让我们尝试通过代码注释逐步解决这个问题
#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);
}
So, basically, it works by going over any given datatype, reinterpreting as bytes, and swapping the bytes. 因此,基本上,它可以通过遍历任何给定的数据类型,将其重新解释为字节并交换字节来工作。
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