C++ 如何从函数返回一个字符数组？
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How do I return a char array from a function?
提问by Datoxalas
I've tried the following:
我尝试了以下方法：
char[10] testfunc()
{
char[10] str;
return str;
}
回答by Xeo
Best as an out parameter:
最好作为输出参数：
void testfunc(char* outStr){
char str[10];
for(int i=0; i < 10; ++i){
outStr[i] = str[i];
}
}
Called with
与
int main(){
char myStr[10];
testfunc(myStr);
// myStr is now filled
}
回答by Marijn van Vliet
You have to realize that char[10]is similar to a char*(see comment by @DarkDust). You are in fact returning a pointer. Now the pointer points to a variable (str) which is destroyed as soon as you exit the function, so the pointer points to... nothing!
您必须意识到这char[10]类似于char*（请参阅@DarkDust 的评论）。你实际上是在返回一个指针。现在指针指向一个变量 ( str)，一旦退出函数，该变量就会被销毁，因此指针指向...什么都没有！
Usually in C, you explicitly allocate memory in this case, which won't be destroyed when the function ends:
通常在 C 中，你在这种情况下显式分配内存，当函数结束时它不会被销毁：
char* testfunc()
{
char* str = malloc(10 * sizeof(char));
return str;
}
Be aware though! The memory pointed at by stris now NEVER destroyed. If you don't take care of this, you get something that is known as a 'memory leak'. Be sure to free()the memory after you are done with it:
不过要注意！指向的内存str现在永远不会被破坏。如果你不注意这一点，你会得到一些被称为“内存泄漏”的东西。free()完成后一定要记住：
foo = testfunc();
// do something with your foo
free(foo);
回答by Flinsch
As you're using C++ you could use std::string.
当您使用 C++ 时，您可以使用std::string.
回答by Felice Pollano
a char array is returned by char*, but the function you wrote does not work because you are returning an automatic variable that disappear when the function exits. Use something like this:
char* 返回一个 char 数组，但您编写的函数不起作用，因为您返回的是一个自动变量，该变量在函数退出时消失。使用这样的东西：
char *testfunc() {
char* arr = malloc(100);
strcpy(arr,"xxxx");
return arr;
}
This of course if you are returning an array in the C sense, not an std:: or boost:: or something else. As noted in the comment section: remember to free the memory from the caller.
当然，如果您返回的是 C 意义上的数组，而不是 std:: 或 boost:: 或其他东西。如评论部分所述：记得从调用者那里释放内存。
回答by MSalters
With Boost:
带升压：
boost::array<char, 10> testfunc()
{
boost::array<char, 10> str;
return str;
}
A normal char[10](or any other array) can't be returned from a function.
char[10]不能从函数返回普通（或任何其他数组）。
回答by devcodexyz
when you create local variables inside function that are created in stack most likely get overwritten in memory when exiting the function. so code like this in most c++ implementations will not work:
当您在函数内部创建在堆栈中创建的局部变量时，很可能在退出函数时在内存中被覆盖。所以在大多数 C++ 实现中这样的代码将不起作用：
char[] pupulateChar()
{
char* ch = "wonet return me";
return ch;
}
a fix is to create the variable that want to be populated outside the function or where you want to use it then pass it as parameter and manipulate function, example:
解决方法是创建要填充到函数外部或要使用它的地方的变量，然后将其作为参数传递并操作函数，例如：
void populateChar(char* ch){
strcpy(ch,"fill me will, this will stay",size); // this will work as long it won overflow it.
}
int main(){
char ch[100]; // reserve memory in stack outside the function
populateChar(ch); //populate array
}
c++11 solution using std::move(ch) to cast lvalues to rvalues
使用 std::move(ch) 将左值转换为右值的 c++11 解决方案
void populateChar(char* && fillme){
fillme = new char[20];
strcpy(fillme, "this worked for me");
}
int main(){
char* ch;
populateChar(std::move(ch));
return 0;
}
or this option in c++11:
或 c++11 中的此选项：
char* populateChar(){
char* ch = "test char";
// will change from lvalue to r value
return std::move(ch);
}
int main(){
char* ch = populateChar();
return 0;
}
