示例

class String {
private:
const char *name;
size_t len;

public:
const char * c_str() {return name;}
};

int main(int argc, char* argv[])
{
String string;

printf("string is '%s'\n", answer);
return 0;
}

解决方案
不要将结构传递给自变量数目可变的函数。

class String {
private:
const char *name;
size_t len;

public:
const char * c_str() {return name;}
};

int main(int argc, char* argv[])
{
String string;

printf("string is '%s'\n", answer.c_str() );
return 0;
}