How to read an integer value in char type?-CodePudding

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?

CodePudding user response：

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);

CodePudding user response：

The warning message you are seeing is indicating that the scanf() function is expecting an int * (a pointer to an integer) as the second argument, but you are providing a char * (a pointer to a character) instead.

To avoid the warning message and ensure that the program works correctly, you should use the correct format specifier for a char variable in the scanf() function.

For example, you can use the %c format specifier to read a single character from the user and store it in a char variable:

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