仅选择字符串C ++中的前几个字符 [英] Selecting only the first few characters in a string C++

问题描述

我想使用C ++选择字符串的前8个字符。现在，我创建一个长度为8个字符的临时字符串，并用另一个字符串的前8个字符填充它。

I want to select the first 8 characters of a string using C++. Right now I create a temporary string which is 8 characters long, and fill it with the first 8 characters of another string.

但是，如果另一个字符串不是8个字符

However, if the other string is not 8 characters long, I am left with unwanted whitespace.

string message = "        ";

const char * word = holder.c_str();

for(int i = 0; i<message.length(); i++)
    message[i] = word[i];

如果单词是 123456789abc ，此代码可以正常工作，并且消息包含 12345678 。

If word is "123456789abc", this code works correctly and message contains "12345678".

但是，如果单词较短，则类似于 1234 ，消息最终显示为 1234

However, if word is shorter, something like "1234", message ends up being "1234 "

如何选择字符串的前八个字符，或者如果整个字符串少于8个字符，则整个字符串？

How can I select either the first eight characters of a string, or the entire string if it is shorter than 8 characters?

推荐答案

只需使用 std :: string :: substr ：

std::string str = "123456789abc";
std::string first_eight = str.substr(0, 8);

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