交换两个字符串指针 [英] Swapping two string pointers

问题描述

我必须为char []在C和我希望他们之间切换，只swaping指针
到数组，而不是一个字符在一个时间，所以我写了这个code：

i have to char[] in C and i wanted to swap between them, by only swaping the pointer to the array and not one char at a time so i wrote this code:

#include <stdio.h>
void fastSwap (char **i, char **d)
{
    char *t = *d;
    *d = *i;
    *i = t;
}
int main ()
{
    char num1[] = "012345678910";
    char num2[] = "abcdefghujk";
    fastSwap ((char**)&num1,(char**)&num2);
    printf ("%s\n",num1);
    printf ("%s\n",num2);
    return 0;
}

我得到这个输出（注意最后4个字符）

I get this output (note the last 4 characters)

abcdefgh8910
01234567ujk

当我想到：

abcdefghujk
012345678910

请注意：我的工作在64位Linux系统

NOTE: I am working on a 64 bit Linux System.

推荐答案

您不能修改 NUM1 和 NUM2 ，您的code应该工作，如果你的测试是不是：

You can't modify the addresses of num1 and num2, your code should work if your test was instead:

int main ()
{
    char num1[] = "012345678910";
    char num2[] = "abcdefghujk";
    char *test1 = num1;
    char *test2 = num2;
    fastSwap (&test1,&test2);
    printf ("%s\n",test1);
    printf ("%s\n",test2);
    return 0;
}

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