在C ++中传递C-Strings的问题

Question

I'm not C++ developer and I'm trying to figure out why when I return a C-string from a function I'm getting garbage out.

#include <stdio.h>

const char* tinker(const char* foo);

int main(void)
{
    const char* foo = "foo";
    foo= tinker(foo);
    printf(foo); // Prints garbage
    return 0;
}

const char* tinker(const char* foo)
{
    std::string bar(foo);
    printf(bar.c_str());  // Prints correctly
    return bar.c_str();
}

Answer 1

You're returning a C-string that's based on the internal memory of a std::string . However, this is being printed AFTER your bar is destructed. This memory is garbage by the time it reaches the printf(foo); line.

Answer 2

You're returning a pointer to a buffer internal to bar , but then you're destroying bar (and the buffer) before you use that pointer.

If you need the content of bar after the function returns, return bar instead of bar.c_str() .

Answer 3

To add to what others said, this is one way to get it to work:

#include <stdio.h>

std::string tinker(const char* foo);
int main(void)
{
     const char* foo = "foo";
     std::string foo2= tinker(foo);
     printf(foo2.c_str()); // Prints correctly
     return 0;
}  

std::string tinker(const char* foo)
{     
     std::string bar(foo);
     return bar; 
} 

Here the std::string object is copied* through the return value and you are printing the copy in main().

_{*popular string implementations employ optimizations like reference counting that avoid actually copying the string. In addition, in C++0x the string object will be moved, not copied.}

Answer 4

Because you're allocating bar in tinker's stack. When tinker ends, its stack is reclaimed, bar is deleted, so the pointer you're returning points to deallocated memory, ie it's not valid anymore when you exit the function.

Answer 5

return bar.c_str();

该行返回临时对象的char* ，该函数在函数末尾被删除，因此main（） foo指向已删除的对象char* 。

Answer 6

Because bar lives on tinker's stack; after leaving the function, its c_str() is gone.