从数组中删除找到空格和标点符号的字符 [英] Remove character from array where spaces and punctuation marks are found

问题描述

在我的程序中，我检查整个cstring，如果发现任何空格或标点符号，只是添加空字符到该位置，但是complilor给我一个错误：空字符常量。

In my program, I am checking whole cstring, if any spaces or punctuation marks are found, just add empty character to that location but the complilor is giving me an error: empty character constant.

请帮助我，在我的循环中我检查这样

Please help me out, in my loop i am checking like this

if(ispunct(str1[start])) {
    str1[start]=''; // << empty character constant. 
}
if(isspace(str1[start])) {
    str1[start]=''; // << empty character constant. 
}

这是我的错误，请纠正我。

This is where my errors are please correct me.

例如， str ,, ing，输出应为 string 。

推荐答案

尝试这个（显式地请求 cstring / p>

Try this (as you asked for cstring explicitly):

char str1[100] = "str,, ing";

if(ispunct(str1[start]) || isspace(str1[start])) {
    strncpy(str1 + start, str1 + start + 1, strlen(str1) - start + 1);
}

好吧，这只是在纯粹 c 语言，有更有效的解决方案（看看@Michael Plotke 的详细答案）。

Well, doing this just in pure c language, there are more efficient solutions (have a look at @MichaelPlotke's answer for details).

但您也明确要求 c ++ ，我建议如下的解决方案：

But as you also explicitly ask for c++, I'd recommend a solution as follows:

注意 可以使用标准c ++算法进行'plain'c-样式字符数组。你只需要将你的谓词条件移除到一个小的帮助函数中，并使用 std :: remove_if（）算法：

Note you can use the standard c++ algorithms for 'plain' c-style character arrays also. You just have to place your predicate conditions for removal into a small helper functor and use it with the std::remove_if() algorithm:

struct is_char_category_in_question {
    bool operator()(const char& c) const;
};

后来使用：

#include <string>
#include <algorithm>
#include <iostream>
#include <cctype>
#include <cstring>

// Best chance to have the predicate elided to be inlined, when writing 
// the functor like this:
struct is_char_category_in_question {
    bool operator()(const char& c) const {
        return std::ispunct(c) || std::isspace(c);
    }
};

int main() {
    static char str1[100] = "str,, ing";
    size_t size = strlen(str1);

    // Using std::remove_if() is likely to provide the best balance from perfor-
    // mance  and code size efficiency you can expect from your compiler 
    // implementation.
    std::remove_if(&str1[0], &str1[size + 1], is_char_category_in_question());

    // Regarding specification of the range definitions end of the above state-
    // ment, note we have to add 1 to the strlen() calculated size, to catch the 
    // closing `\0` character of the c-style string being copied correctly and
    // terminate the result as well!

    std::cout << str1 << endl; // Prints: string
}

查看这个可编译的工作示例此处。

See this compilable and working sample also here.

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