如何在 C++ 中将字符串转换为字符数组？
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How to convert string to char array in C++?
提问by Brian Brown
I would like to convert stringto chararray but not char*. I know how to convert string to char*(by using mallocor the way I posted it in my code) - but that's not what I want. I simply want to convert stringto char[size]array. Is it possible?
我想转换string为char数组但不是char*. 我知道如何将字符串转换为char*（通过使用malloc或我在代码中发布它的方式） - 但这不是我想要的。我只是想转换string为char[size]数组。是否可以？
#include <iostream>
#include <string>
#include <stdio.h>
using namespace std;
int main()
{
// char to string
char tab[4];
tab[0] = 'c';
tab[1] = 'a';
tab[2] = 't';
tab[3] = 'string temp = "cat";
char tab2[1024];
strcpy(tab2, temp.c_str());
';
string tmp(tab);
cout << tmp << "\n";
// string to char* - but thats not what I want
char *c = const_cast<char*>(tmp.c_str());
cout << c << "\n";
//string to char
char tab2[1024];
// ?
return 0;
}
回答by Chowlett
Simplest way I can think of doing it is:
我能想到的最简单的方法是：
string temp = "cat";
char tab2[1024];
strncpy(tab2, temp.c_str(), sizeof(tab2));
tab2[sizeof(tab2) - 1] = 0;
For safety, you might prefer:
为了安全起见，您可能更喜欢：
string temp = "cat";
char * tab2 = new char [temp.length()+1];
strcpy (tab2, temp.c_str());
or could be in this fashion:
或者可以采用这种方式：
const char *array = tmp.c_str();
回答by rad
Ok, i am shocked that no one really gave a good answer, now my turn. There are two cases;
好吧，我很震惊，没有人真正给出好的答案，现在轮到我了。有两种情况；
A constant char arrayis good enough for you so you go with,
char *array = &tmp[0];Or you need to modifythe char array so constant is not ok, then just go with this
const char *array = tmp.c_str();
一个常量字符数组对你来说已经足够了，所以你可以使用，
char *array = &tmp[0];或者你需要修改char 数组所以常量不行，然后就用这个
std::string myWord = "myWord"; char myArray[myWord.size()+1];//as 1 char space for null is also required strcpy(myArray, myWord.c_str());
Both of them are just assignment operationsand most of the time that is just what you need, if you really need a new copy then follow other fellows answers.
它们都只是分配操作，大多数时候这正是您所需要的，如果您确实需要一个新副本，请遵循其他人的答案。
回答by rad
Easiest way to do it would be this
最简单的方法是这样
str.copy(cstr, str.length()+1); // since C++11
cstr[str.copy(cstr, str.length())] = 'string line="hello world";
char * data = new char[line.size() + 1];
copy(line.begin(), line.end(), data);
data[line.size()] = 'strcpy(tab2, tmp.c_str());
';
'; // before C++11
cstr[str.copy(cstr, sizeof(cstr)-1)] = 'auto tab2 = std::make_unique<char[]>(temp.size() + 1);
std::strcpy(tab2.get(), temp.c_str());
'; // before C++11 (safe)
回答by Youka
string n;
cin>> n;
char b[200];
for (int i = 0; i < sizeof(n); i++)
{
b[i] = n[i];
cout<< b[i]<< " ";
}
It's a better practice to avoid C in C++, so std::string::copyshould be the choice instead of strcpy.
在 C++ 中避免使用 C 是一种更好的做法，因此应该选择std::string::copy而不是strcpy。
回答by David Schwartz
Just copy the string into the array with strcpy.
只需将字符串复制到数组中strcpy。
回答by Tharindu Dhanushka
Try this way it should be work.
试试这种方式它应该是有效的。
##代码##回答by Marrow Gnawer
Try strcpy(), but as Fred said, this is C++, not C
尝试 strcpy()，但正如 Fred 所说，这是 C++，而不是 C
回答by Fred Larson
You could use strcpy(), like so:
你可以使用strcpy()，像这样：
Watch out for buffer overflow.
注意缓冲区溢出。
回答by emlai
If you don't know the size of the string beforehand, you can dynamically allocate an array:
如果事先不知道字符串的大小，可以动态分配一个数组：
##代码##回答by SquircKle
Well I know this maybe rather dumb thanand simple, but I think it should work:
好吧，我知道这可能比简单更愚蠢，但我认为它应该有效：
