problem with atoi ()

Hi,

I have a problem with atoi ()
My code is as follows:

Code:

# Include <stdlib.h> 
# Include <stdio.h> 
int main (void) ( 
    char* x = "f"; 
    int command; 
    command = atoi (x); 
    printf ("number:% d", command); 
   return 0; 
)

How to remove the link, atoi () returns 0 if could not be converted to char. with me, the program always 0 Why?

 

because "f" is not a number. as simple as that

 

what is a number "f" be? 1? 3? 19?
if you work instead of "f" as "650" uses it.

btw:
printf is not c, cpp court and in 'real'?

 

I probably hit the wrong function. in reality I would like to determine the ANSI code given by a letter that is then returned. I found the function in CharToInt R.oo, but I'd like something not so large scale. Is there something?

 

What do you want exactly? What number should I get out at "f"?

 

The ANSI code in which case, 69 decimal or 45 hex

 

you can use

Code:

printf ("Number:% d", 'f');

 

Thank you. But why, I tap in the dark and always think so complicated and it is so easy.

 

since you are indeed benefit:

why does it not here?

Code:

char* book = "x"; 
int test = (int) book;     
court <<"number: "<<test <<endl;

It always returns 4,198,960th Why?

 

that is the memory address of the chars

 

nt test = (int) * book;

 

I am now somewhat confused. of course it must be int (* char be). Thank you

Does all the other way around? So in some book = char (test)? So it is not hot but how do this so that it's going?

 

Code:

int test = 120; 
char book = test;

use it.

 

I am not sure if you have your answer or not but the function you are using takes character and not string.