How to read an integer value in char type?

Question

I wanted to make a little experiment with scanf(). I wanted to read a small (<=255) integer from the user and store it in a char type.

I did:

char ch;
scanf("%d",&ch);

It works, but I want to satisfy the compiler and not to get this warning:

warning: format specifies type 'int *'
but the argument has type 'char *' [-Wformat]
scanf("%d",&ch);

Any idea?

`%d` means you are expecting an integer. Use `scanf("%c",&ch)` instead. — Fractalism, Jan 23 '23 at 11:15
As Fractalism already commented, `%c` is used for `char` type. You can check [this website](https://en.wikibooks.org/wiki/C_Programming/Simple_input_and_output#Placeholder_codes) for the basics C place holders. — Parker, Jan 23 '23 at 11:18
But OP does not want to read a character. For example wants to read `255` as a small integer (3 characters). — Weather Vane, Jan 23 '23 at 11:19
To read the range 0..255 you need `unsigned char c; scanf("%hhu", &c);` — Weather Vane, Jan 23 '23 at 11:24

score 3 · Answer 1 · answered Jan 23 '23 at 11:31

First of all, to store a value 0 - 255, you should never use char. In fact you should never use char for the intention of storing anything but characters and strings. This is because that type has implementation-defined signedness: Is char signed or unsigned by default?

You could do:

unsigned char val;
scanf("%hhu", &val);

Although it is often best practice to use the portable types from stdint.h:

#include <stdint.h>
#include <inttypes.h>  // (actually includes stdint.h internally)

uint8_t val; 
scanf("%"SCNu8, &val);

How to read an integer value in char type?

1 Answers1