c ++将char数组传递给函数并进行更改

Question

I try to make a function that can change the content of a specified char array

void change_array(char *target)
{
    target="hi";
}

int main()
{
    char *a[2];
    change_array(a[1]);
    cout<<*(a[1]);
}

But then the content of a[1] stays at 0x0(void)

Answer 1

First, your function has a copy of the pointer passed to it, so there is no effect seen on the caller side. If you want to modify the function argument, pass a reference:

void change_array(char*& target) { ... }
//                     ^

Second, you cannot/should not bind a non-const pointer to a string literal. Use const char* instead.

void change_array(const char*& target) { ... }
//                ^^^^^      ^

int main()
{
    const char* a[2];
    change_array(a[1]);
    cout<<*(a[1]);
}

Answer 2

When you pass an argument to a function, it's normally passed by value, meaning its value is copied. If you want to change it you have to pass it by reference. The same goes for pointers, if you want to change a pointer then you need to pass it by reference as well:

void change_array(const char*& target) { ... }

Answer 3

You need to pass it as a reference:

void change_array(char*&target)
{
    target="hi";
}

Otherwise, you will just change the local copy of target , which won't make any difference to the value outside of the function.

Answer 4

Try this design instead:

std::string get_string()
{
    return "hi";
}

int main()
{
    std::string a[2];
    a[1] = get_string();
    std::cout<< a[1];
}

Salient points:

• return by value
• use std::string