如何设置char *值从std字符串(c_str())不工作 [英] how to set char * value from std string (c_str()) not working
问题描述
我不知道,但这不是我的工作我获取garbege值,当我尝试设置char *值从返回std字符串的函数:
i dont know but this not working for me im getting garbege value when i try to set char * value from function that returns std string :
string foo()
{
string tmp ="dummy value";
return tmp;
}
char* cc = (char *) foo().c_str(); // if i remove the casting im getting error
// when i print the cc i get garbage
printf("%s",cc);
推荐答案
<$ cc指向的数据的生命周期$ c> cc 与字符串的生命周期相同(最多 - 如果修改字符串,它甚至更短)。
The lifetime of the data pointed to by cc
is the same as the lifetime of the string it came from (at best - if you modify the string it's even shorter).
在你的情况下, foo()
的返回值是在 cc $ c的初始化结束时销毁的临时$ c>。
In your case, the return value of foo()
is a temporary that is destroyed at the end of the initialization of cc
.
为避免 char * cc = foo()。c_str()
不应转换为 char *
,应切换到 const char * cc
,因为 const char *
是什么 c_str()
返回。但
To avoid the compilation error in char *cc = foo().c_str()
you shouldn't cast to char*
, you should switch to const char *cc
, since const char*
is what c_str()
returns. That still doesn't solve the main problem, though.
最简单的修复是:
printf("%s", foo().c_str()); // if you don't need the value again later
const string s = foo();
const char *cc = s.c_str(); // if you really want the pointer - since it's
// in the same scope as s, and s is const,
// the data lives as long as cc's in scope.
string s = foo();
printf("%s", s.c_str()); // if you don't store the pointer,
// you don't have to worry about it.
std::cout << foo(); // printf isn't bringing much to this party anyway.
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