如何从C中的字符串中删除所有出现的给定字符？

I'm attempting to remove a character from a string in C. The problem I am having with my code is that it removes the first instance of the character from the string but also wipes everything after that character in the string too. For example, removing 'l' from 'hello' prints 'he' rather than 'heo'

我正在尝试从C中的字符串中删除一个字符。我的代码问题是它从字符串中删除了字符的第一个实例，但也擦除了字符串中该字符后的所有内容。例如，从'hello'中删除'l'会打印'he'而不是'heo'

int i;
char str1[30] = "Hello", *ptr1, c = 'l';
ptr1 = str1;
for (i=0; i<strlen(str1); i++)
{
    if (*ptr1 == c) *ptr1 = 0;
    printf("%c\n", *ptr1);
    ptr1++;
}

I need to use pointers for this and would like to keep it as simple as possible since I'm a beginner in C. Thanks

我需要使用指针，并希望尽可能简单，因为我是C的初学者。谢谢

7 个解决方案

#1

You can do it like this:

你可以这样做：

void remove_all_chars(char* str, char c) {
    char *pr = str, *pw = str;
    while (*pr) {
        *pw = *pr++;
        pw += (*pw != c);
    }
    *pw = '\0';
}

int main() {
    char str[] = "llHello, world!ll";
    remove_all_chars(str, 'l');
    printf("'%s'\n", str);
    return 0;
}

The idea is to keep a separate read and write pointers (pr for reading and pw for writing), always advance the reading pointer, and advance the writing pointer only when it's not pointing to a given character.

我们的想法是保持一个单独的读写指针（pr用于读取和pw用于写入），始终使读取指针前进，并且仅在写入指针未指向给定字符时才前进。

#2

If you remove the characters in place you will have to shift the rest of the string one place to the left every time you remove a character, this is not very efficient. The best way is to have a second array that takes the filtered string. For example you can change your code like this.

如果删除了字符，则每次删除字符时都必须将字符串的其余部分向左移动一个位置，这不是很有效。最好的方法是使用第二个数组来获取过滤后的字符串。例如，您可以像这样更改代码。

int i;
char str1[30] = "Hello", *ptr1, c = 'l';
char str2[30] = {0}, *ptr2;
ptr1 = str1;
ptr2 = str2;
for (i=0; i<strlen(str1); i++)
{
    if (*ptr1 != c) *ptr2++=*ptr1;
    ptr1++;
}
printf("%s\n", str2);

#3

char str1[30] = "Hello", *prt1, c = 'l';
char str2[30], *prt2;
prt1 = str1;
prt2 = str2;
while(*prt1 != 0)
{
    if(*prt1 != c)
    {
         *prt2 = *prt1;
         prt2++;  
    }
    prt1++;
}
*prt2 = '\0';

#4

The problem is that when you encounter the first character that matches c, you insert a null character right there. That means you're essentially cutting off the rest of the string.

问题是当你遇到匹配c的第一个字符时，你在那里插入一个空字符。这意味着你基本上切断了其余的字符串。

What you need to do is when you find a matching character, move the following characters back one position. Then you need to insert the null character at the very end depending on how many characters you have removed.

您需要做的是找到匹配的字符，将后面的字符移回一个位置。然后，您需要在最后插入空字符，具体取决于您删除的字符数。

#5

i know that it is a type of duplicate answer, but this code is function's version for solving the problem. I thought that as the questioner was a beginner, he might learn much from decomposed version of problem.

我知道这是一种重复的答案，但这段代码是解决问题的函数版本。我认为，由于提问者是初学者，他可能会从分解版本的问题中学到很多东西。

int del_x_char(char *p, int x)
{
    char *q;
    x=first_occurance(p, 'i')/*you can replace any character that you want delete with 'i'*/
    q=p+x;
    while(*q=*(q+1))
        q++;
    *q='\0';
    return 0;
}
int first_occurance(char *q, char phar)
{
    int i=0;
    while(*q)
    {   
        if(*q++==phar)
            return i;
        i++;
    }
    return -1;
}

#6

just change

只是改变

if (*ptr1 == c) *ptr1 = 0;

至

if (*ptr1 == c) continue;

as @ouah said, it breaks at the first NULL character..

正如@ouah所说，它在第一个NULL字符处断开..

#7

-1

C defines a string as "a contiguous sequence of characters terminated by and including the first null character"

C将字符串定义为“由第一个空字符终止并包含第一个空字符的连续字符序列”

#1