Narrowing conversion from 'long' to signed type 'char' is implementation-defined (strtol function in C)

Question

I'm trying to convert an inputted character to an integer by using strtol. Here's part of the source code:

char option;
char *endptr;

printf("=========================================Login or Create Account=========================================\n\n");
while(1) {
    printf("Welcome to the Bank management program! Would you like to 1. Create Account or 2. Login?\n>>> ");
    fgets(&option, 1, stdin);
    cleanStdinBuffer();
    option = strtol(&option, &endptr, 10);

In the strtol function, I'm getting a warning saying:

Clang-Tidy: Narrowing conversion from 'long' to signed type 'char' is implementation-defined

Can someone tell me what I'm doing wrong?

Answer 1

Clang-Tidy is warning you about the implicit conversion you are doing here where you are assign the long return value of strtol to a char:

option = strtol(&option, &endptr, 10);

If this is intentional and you are sure the value will be in the [-128,127] range that isn't necessarily an issue (it's just a warning), but even then I would advice to explicitly cast the return-type of strtol, use int8_t instead of char and not reuse the option variable for the return value. In other words:

int8_t value = (int8_t)strtol(&option, &endptr, 10);

If it wasn't intentional I would recommend you to simply use long as type for the variable you assign the return value of strtol, so:

long value = strtol(&option, &endptr, 10);

What Clang-tidy doesn't warn you about is that the first argument to strtol should be a pointer to a char buffer containing a 0-terminated string, not a pointer to a single char. This is also an issue for fgets. There are two ways to solve this, either:

1. Make option a char array of at least two chars,

2. Use fgetc instead and modify your code into something like this:

   int option = fgetc(stdin);
   if (option == '1') {
       /*Create Account */
   } else if (option == '2') {
       /* Login */
   }
   else {
       /* Error */
   }
   

I think the latter looks much cleaner.

Answer 2

char can only hold a very little subset of the long values. strtol returns long and you assign it to char.

Answer 3

This call of fgets

fgets(&option, 1, stdin);

always sets the character option to the terminating zero character '\0' provided that the user did not interrupt the input.

Here is a demonstrative program.

#include <stdio.h>

int main(void) 
{
    char c = 'A';
    
    printf( "Before calling fgets c = %d\n", c );
    
    fgets( &c, 1, stdin );
    
    printf( "After  calling fgets c = %d\n", c );

    return 0;
}

The program output is

Before calling fgets c = 65
After  calling fgets c = 0

independent on what the user will enter. Here is the value 65 is the ASCII code of the character 'A' that was stored in the variable c before calling fgets.

If you want to enter a character then you should use

scanf( " %c", &input );

Pay attention to the blank before the conversion specifier.

After this call you can check whether the user typed a digit like

#include <ctypes.h>

//...

if ( isdigit( ( unsigned char )input ) ) input = input - '0';

and set the variable input to the corresponding integer value in the range [0, 9].