C ++和printf-奇怪的字符输出

Question

I'm a complete newb to C++, but not to Java, C#, JavaScript, VB. I'm working with a default C++ console app from Visual Studio 2010.

In trying to do a printf I get some strange characters. Not the same each time which tells me they may be looking at different memory location each time I run it.

Code:

#include "stdafx.h"
#include <string>

using namespace std;

class Person
{
public:
    string first_name;
};

int _tmain(int argc, _TCHAR* argv[])
{
    char somechar;
    Person p;
    p.first_name = "Bruno";

    printf("Hello %s", p.first_name);
    scanf("%c",&somechar);
    return 0;
}

Answer 1

The problem is that printf / scanf are not typesafe. You're supplying a std::string object where printf expects a const char* .

One way to fix this is to write

printf("Hello %s", p.first_name.c_str());

However, since you're coding in C++, it's a good idea to use I/O streams in preference to printf / scanf :

std::cout << p.first_name << std::endl;
std::cin >> c;

Answer 2

Convert the string to a c-string.

printf("Hello %s", p.first_name.c_str()); 

Also, since you're using C++, you should learn about cout as opposed to printf!

Answer 3

Use printf("Hello %s",p.first_name.c_str()); !

Answer 4

printf("Hello %s", p.first_name.c_str());

但是，如果使用的是c ++，为什么不使用iostream？

Answer 5

You cannot pass C++ std::string objects into printf . printf only understands the primitive types like int , float , and char* . Your compiler should be giving you a warning there; if it's not, crank up your warning level.

Since you're using C++, you really should be using std::cout for text output, and that does understand std::string objects. If you really have to use printf for some reason, then convert the std::string to a const char* by calling the c_str() method on it.

Answer 6

printf("%s") accepts a c-style string which is terminated by a '\\0' . However, string object is C++ object which is different from a c-style string. You should use std::cout which is overloaded to handle string type directly, as shown below.

std::cout << p.first_name;