Question

此代码用%20替换了空格，char数组应该包含足够的空格来添加%20。

你能帮助我理解为什么我会在*str = *tmp; // segfault at this location..得到段错误吗？

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


char* encodeSpace(char* str) {
    char *orig = str;
    while (*str++);

     str--;
     char *tmp = --str;
     while (*tmp-- == ' ');
     tmp++;
     while (tmp != orig) {
        while(*tmp != ' ' && tmp != orig) {

        *str = *tmp; // segfault at this location.. 
         tmp--;
         str--;
        }
        *str-- = '0';
        *str-- = '2';
        *str-- = '%';
        tmp--;
     }
     return tmp;
}

main()
{
   printf("output is %s", encodeSpace("My Name is john         "));
}

Answer 1

初始主要问题

您遇到分段错误，因为您尝试修改字符串文字（函数的参数）。

字符串文字通常位于只读内存中，可能无法合法修改。尝试这样做会导致未定义的行为。分段错误是一种合法的未定义行为。

如果你写的话，你会没事的。

int main(void)
{
    char str[] = "My Name is john         ";
    printf("output is %s", encodeSpace(str));
    return 0;
}

或甚至（使用C99复合文字）：

int main(void)
{
    printf("output is %s", encodeSpace((char[]){"My Name is john         "}));
    return 0;
}

算法问题

因为你可以在一个循环中递减tmp两次，所以你不能保证它何时到达orig指针。此代码在assert()：

中崩溃

#include <assert.h>
#include <inttypes.h>
#include <stdio.h>
#include <string.h>

static
char *encodeSpace(char *str)
{
    char *orig = str;
    printf("-->> str = %p <<%s>>\n", (void *)str, str);

    while (*str++)
        ;

    str--;
    char *tmp = --str;
    while (*tmp-- == ' ')
        ;
    tmp++;
    printf("tmp = %p <<%s>>\n", (void *)tmp, tmp);
    printf("str = %p <<%s>>\n", (void *)str, str);
    while (tmp != orig)
    {
        while (*tmp != ' ' && tmp != orig)
        {
            printf("--1: tmp = %p <<%s>>; str = %p <<%s>>\n", (void *)tmp, tmp, (void *)str, str);
            *str = *tmp; // segfault at this location..
            tmp--;
            str--;
            printf("--2: tmp = %p <<%s>>; str = %p <<%s>>\n", (void *)tmp, tmp, (void *)str, str);
        }
        //if (tmp != orig)
        //{
        *str-- = '0';
        *str-- = '2';
        *str-- = '%';
        tmp--;
        //}
        printf("--3: tmp = %p <<%s>>; str = %p <<%s>>\n", (void *)tmp, tmp, (void *)str, str);
        assert(tmp >= orig);
    }
    printf("<<-- tmp = <<%s>>\n", tmp);
    return tmp;
}

int main(void)
{
    printf("output is <<%s>>\n", encodeSpace((char[]){"My name is John         "}));
    return 0;
}

删除注释，它不再崩溃（tmp的第二次减少受到保护）。但输出并不是你想要的。

-->> str = 0x7fff54cb44c0 <<My name is John         >>
tmp = 0x7fff54cb44ce <<n         >>
str = 0x7fff54cb44d7 << >>
--1: tmp = 0x7fff54cb44ce <<n         >>; str = 0x7fff54cb44d7 << >>
--2: tmp = 0x7fff54cb44cd <<hn        n>>; str = 0x7fff54cb44d6 << n>>
--1: tmp = 0x7fff54cb44cd <<hn        n>>; str = 0x7fff54cb44d6 << n>>
--2: tmp = 0x7fff54cb44cc <<ohn       hn>>; str = 0x7fff54cb44d5 << hn>>
--1: tmp = 0x7fff54cb44cc <<ohn       hn>>; str = 0x7fff54cb44d5 << hn>>
--2: tmp = 0x7fff54cb44cb <<John      ohn>>; str = 0x7fff54cb44d4 << ohn>>
--1: tmp = 0x7fff54cb44cb <<John      ohn>>; str = 0x7fff54cb44d4 << ohn>>
--2: tmp = 0x7fff54cb44ca << John     John>>; str = 0x7fff54cb44d3 << John>>
--3: tmp = 0x7fff54cb44c9 <<s John  %20John>>; str = 0x7fff54cb44d0 << %20John>>
--1: tmp = 0x7fff54cb44c9 <<s John  %20John>>; str = 0x7fff54cb44d0 << %20John>>
--2: tmp = 0x7fff54cb44c8 <<is John s%20John>>; str = 0x7fff54cb44cf << s%20John>>
--1: tmp = 0x7fff54cb44c8 <<is John s%20John>>; str = 0x7fff54cb44cf << s%20John>>
--2: tmp = 0x7fff54cb44c7 << is Johnis%20John>>; str = 0x7fff54cb44ce <<nis%20John>>
--3: tmp = 0x7fff54cb44c6 <<e is J%20is%20John>>; str = 0x7fff54cb44cb <<J%20is%20John>>
--1: tmp = 0x7fff54cb44c6 <<e is J%20is%20John>>; str = 0x7fff54cb44cb <<J%20is%20John>>
--2: tmp = 0x7fff54cb44c5 <<me is e%20is%20John>>; str = 0x7fff54cb44ca << e%20is%20John>>
--1: tmp = 0x7fff54cb44c5 <<me is e%20is%20John>>; str = 0x7fff54cb44ca << e%20is%20John>>
--2: tmp = 0x7fff54cb44c4 <<ame isme%20is%20John>>; str = 0x7fff54cb44c9 <<sme%20is%20John>>
--1: tmp = 0x7fff54cb44c4 <<ame isme%20is%20John>>; str = 0x7fff54cb44c9 <<sme%20is%20John>>
--2: tmp = 0x7fff54cb44c3 <<name iame%20is%20John>>; str = 0x7fff54cb44c8 <<iame%20is%20John>>
--1: tmp = 0x7fff54cb44c3 <<name iame%20is%20John>>; str = 0x7fff54cb44c8 <<iame%20is%20John>>
--2: tmp = 0x7fff54cb44c2 << name name%20is%20John>>; str = 0x7fff54cb44c7 << name%20is%20John>>
--3: tmp = 0x7fff54cb44c1 <<y na%20name%20is%20John>>; str = 0x7fff54cb44c4 <<a%20name%20is%20John>>
--1: tmp = 0x7fff54cb44c1 <<y na%20name%20is%20John>>; str = 0x7fff54cb44c4 <<a%20name%20is%20John>>
--2: tmp = 0x7fff54cb44c0 <<My ny%20name%20is%20John>>; str = 0x7fff54cb44c3 <<ny%20name%20is%20John>>
--3: tmp = 0x7fff54cb44c0 <<My ny%20name%20is%20John>>; str = 0x7fff54cb44c3 <<ny%20name%20is%20John>>
<<-- tmp = <<My ny%20name%20is%20John>>
output is <<My ny%20name%20is%20John>>

我认为您的算法很复杂，因为您正在尝试进行原位更改。我无法解决如何解决它。传入一个输入字符串（可能是一个字符串文字，并且不需要空白填充）和一个输出缓冲区会更简单，你可以在调用中指定它的大小：

#include <stdio.h>

static int encodeSpace(char const *src, char *buffer, size_t buflen)
{
    char *end = buffer + buflen;
    char c;
    char *dst = buffer;

    while ((c = *src++) != '\0' && dst < end)
    {
        if (c != ' ')
            *dst++ = c;
        else if (dst < end - 3)
        {
            *dst++ = '%';
            *dst++ = '2';
            *dst++ = '0';
        }
        else
            return -1;
    }
    *dst = '\0';
    return 0;
}

int main(void)
{
    char buffer[64];

    if (encodeSpace("My name is John", buffer, sizeof(buffer)) == 0)
        printf("output is <<%s>>\n", buffer);

    return 0;
}

输出：

output is <<My%20name%20is%20John>>

Answer 2

@Jonathan Leffler的解决方案非常完美。

我试图修复原始算法，但在原始字符缓冲区中看起来很难更改因为%20占用三个字符位置，如果space字符不够，代码仍然出错。< / p> 如果空间足够，

代码跟随效果很好。

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


char* encodeSpace(char* str) {
    char *orig = str;
    while (*str++);

     str--;
     char *tmp = --str;
     while (*tmp == ' ') tmp--;
     tmp++;
     while (tmp != orig) {
        while(*tmp != ' ') {
         *str = *tmp;
         str--;
         if (tmp == orig)
            break;
         else
           tmp--;
        }
        if (*tmp == ' ') {
          *str-- = '0';
          *str-- = '2';
          *str-- = '%';
          while (*tmp == ' ') tmp--;
        } 
     }
     return ++str;
}

int main()
{

    char str[] = "My Name is john            ";
   printf("output is %s\n", encodeSpace(str));
   return 0;
}

当我尝试复制* str = * tmp时，为什么会发生分段错误

2 个答案:

初始主要问题

算法问题