C++中char[]的子串
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/15586586/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Substring of char[] in c++
提问by Yoda
I have
我有
char t[200];
cin.get(s, 200);
int i = 5;
int j = 10;
Is there any simple way to get substriing(i,j)from tbeside copying every element seperately to the another array? No stringsetc. just char t[200].
有没有简单的方法来获得substriing(i,j)从t每个元素seperately复制到另一个阵列旁边？没有strings等只是char t[200]。
采纳答案by NPE
If you are allowed to modify t, you could set t[j]to 0and then use t + ito get the substring.
如果你被允许修改t，你可以设置t[j]为0然后使用t + i来获取子字符串。
If not, you are going to have to make a copy.
如果没有，您将不得不复制。
That said, why can't you just use std::stringand save yourself the headache?
也就是说，你为什么不能直接使用std::string并避免头痛呢？
回答by Jean-Baptiste Yunès
If you need only to read the data then t+i is what you want, alas you'll have to manage the length of your substring...
如果您只需要读取数据，那么 t+i 就是您想要的，唉，您必须管理子字符串的长度...
char *sub = t+i;
int len = j-i;
printf("%.*s\n",len,sub);
If you need to have a distinct copy of the substring, then you must copy.
如果您需要具有子字符串的不同副本，则必须进行复制。
回答by Mayhem
This should work fine:
这应该可以正常工作：
#include <string.h>
#include <stdlib.h>
#include <iostream>
using namespace std;
int main()
{
char t[200];
cin.get(t, 200);
int i = 5;
int j = 10;
char *to = (char*) malloc(j-i+1);
strncpy(to, t+i, j-i);
to[j-i]='char* to = new char[j-i+1];
';
cout << to;
}
You can use newinstead of malloclike this:
您可以使用new而不是malloc这样：
char* substr(char* arr, int begin, int len)
{
char* res = new char[len];
for (int i = 0; i < len; i++)
res[i] = *(arr + begin + i);
res[len] = 0;
return res;
}
回答by spynet41
char const * beg = t+i;
char const * end = t+j+1;
std::cout.write(beg, end-beg);
回答by Benjamin Lindley
Use two pointers to denote a range within the string.
使用两个指针来表示字符串内的一个范围。
llvm::StringRef sref(t+i, j+1-i);
std:cout << sref;
Or you can use a class that encapsulates this idea. There is something like thatbeing proposed for the standard library. In the meantime, you can write your own, or you can use one from a library. For example, llvm::StringRef.
或者你可以使用一个封装了这个想法的类。有这样的事情正在提议的标准库。同时，您可以自己编写，也可以使用库中的一个。例如，llvm::StringRef。
char newt[200];
// copy j-i chars from position t+i to newt array
strncpy(newt, t + i, j-i);
// now null terminate
newt[j-i] = 0;
回答by suspectus
This does not do any bounds checking to ensure the destination array is large enough
这不会进行任何边界检查以确保目标数组足够大
##代码##