dynamic char array size

[technologist]

Hi,

I want to make char mystring the size of input. Compiler is claiming it can't do this without an explicit number. I don't understand my the formatting is the way it is.

1 2 3 4

char mystring[]; tried as well cout<<endl<<"Take a card ?"; char mystring[10];

cin>>mystring; cout<<mystring; if (mystring[0] == 'y') {cout<<"-y-";}

output:
..\src\main.cpp:183:16: error: storage size of 'mystring' isn't known

[Verideth]

If I'm correct you can only std::cin with strings.

[Peter87]

Why don't you use a std::string instead? Then you would not have to know the size of the string when you create it.

http://www.cplusplus.com/reference/string/string/

Your output is messed up because when you put --- inside a code block it's split up into code and output sections.

[technologist]

...or you can scratch through the array

cout<<endl<<"Draw or Hold ?";
	char io [256];
	cin>>io;
	char * n_ptr = &io[0];
		cout<<"1st character<<"<<*n_ptr;
		cout<<*(n_ptr+1);

output: Dr if you put in Draw

Neat to manipulate the string as pointer array

[technologist]

Or as Peter said use std::string instead:

#include<iostream>
#include<string>

int main ()
{
	std::string str_input;
	std::cin>>str_input;
	std::cout<<str_input[0];
}