如何更改字符串数组中的单个字符？ [英] how to change single char in string array?

问题描述

具有以下内容：

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

char *up(char *);

int main() {
    char initstr[20];
    printf("enter string\n");
    fgets(initstr, 20, stdin);

    char *str = up(initstr);
    printf("%s\n", str);
}

char *up(char *in) {
    char *ret;
    for (ret = in;
         *in != '\n'; 
         *(ret++) = toupper(*(in++))
        );
    return ret;
}

运行方式为：

$./a.out
enter string
abc

#only new line from `printf("%s\n",str);`

来自调试器

Hardware watchpoint 3: in

Old value = 0x7fffffffdc20 "abc\n"
New value = 0x7fffffffdc21 "bc\n"

Hardware watchpoint 2: ret

Old value = 0x7fffffffdc20 "abc\n"
New value = 0x7fffffffdc21 "bc\n"

Hardware watchpoint 3: in

Old value = 0x7fffffffdc21 "bc\n"
New value = 0x7fffffffdc22 "c\n"

Hardware watchpoint 2: ret

Old value = 0x7fffffffdc21 "bc\n"
New value = 0x7fffffffdc22 "c\n"
...

我看到两个变量都在减少，但是我想更改 ret inline ，一个字符一个字符。但是最后（循环之后）， ret 减少为零，并且程序将仅输出 \n 。那么如何在循环头中实现这一目标呢？

I can see that both variables are reducing, but I wanted to change the ret inline, char by char. But at the end (after loop), the ret is reduced to nothing, and the program will only output \n. So how can I achieve this in the loop head?

编辑：
感谢下面的回答，请记住我必须返回指针的第一个地址，可以通过以下方式实现loop_head-only功能：

Thanks to answer below, having in mind I have to return first address of pointer, I can implement loop_head-only function by this:

char *up(char *in){
    char *ret;
    size_t size=strlen(in);
    for(ret=in;
         *in!='\n'; 
         *(ret++)=toupper(*(in++))
        );
    return (ret-size+1);
}

推荐答案

<$ c中的错误$ c> up 是您将 ret 一直增加到换行符（ \n ）并返回 ret 指向字符串中的此字符。相反，您应该返回指向初始字符的指针。

The bug in up is you increment ret all the way to the newline (\n) and return ret pointing to this character in the string. You should instead return a pointer to the initial character.

• 使用索引编写此函数更简单。


• 将所有逻辑打包到 for 子句中时，其内容为空，很难读取且容易出错。


• 注意另外，该字符串可能不包含换行符，因此在空终止符处停止比较安全， toupper（）不会更改换行符。


• 最后，您不应将 char 值传递给 toupper（），因为此函数和所有函数< ctype.h> 中仅定义了类型为 unsigned char 和特殊负值<$ c的值$ c> EOF 。在默认情况下对 char 进行签名的平台上，字符串可能包含负的 char 值，当传递给时可能导致未定义的行为 toupper（）。将它们强制转换为（未签名字符），以避免出现此问题。

• It is simpler to write this function using an index.
• packing all the logic into the for clauses with an empty body is hard to read and error prone.
• Note also that the string might not contain a newline, so it is safer to stop at the null terminator, the newline will not be changed by toupper().
• Finally, you should not pass char values to toupper() because this function and all functions from <ctype.h> is only defined for values of type unsigned char and the special negative value EOF. On platforms where char is signed by default, the string might contain negative char values which may cause undefined behavior when passed to toupper(). Cast these as (unsigned char) to avoid this issue.

修改后的版本：

#include <ctype.h>
#include <stdio.h>

char *up(char *s) {
    for (size_t i = 0; s[i] != '\0'; i++) {
         s[i] = toupper((unsigned char)s[i]);
    }
    return s;
}

int main() {
    char initstr[20];
    printf("enter string\n");
    if (fgets(initstr, sizeof initstr, stdin)) {
        char *str = up(initstr);
        printf("%s\n", str);
    }
    return 0;
}

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