Why does the smallest int, −2147483648, have type 'long'?

Question

For a school project, I've to code the C function printf. Things are going pretty well, but there is one question I can't find a good answer to, so here I am. printf("PRINTF(d) \ : %d\\n", -2147483648);tells me (`gcc -Werror -Wextra -Wall`): error: format specifies type 'int' but the argument has type 'long' [-Werror,-Wformat] printf("PRINTF(d) \ : %d\\n", -2147483648); ~~ ^~~~~~~~~~~ %ldBut if I use an int variable, everything is going well: int i; i = -2147483648; printf("%d", i);Why?EDIT:------I understood many points, and they were very interesting. Anyway, I guess `printf` is using the `<stdarg.h>` librairy and so, `va_arg(va_list ap, type)` should also return the right type. For `%d` and `%i`, obviously the type returned is an `int`. Does it change anything?

Answer 1

The range of a signed int goes from 32,767 to -32,767; what you are trying to print is a long value. The system knows this and because you have turned up the warning levels, it considers that an error. You're saying you want to print an integer, but instead you're trying to print a long.

You can find the range of the types in <limits.h> for c, <climits> for c++