Command Line Arguments and Chars - c

Question

What is the proper way to receive a command line argument from the user and compare it to a single char? For example if the user typed "Y" for yes run some function and "N" for no would run another function.
My main error is "comparison between ptr and integer" & "too few arguments to function call, expected 3 have 1"
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void calculate();
void verify();
int main (int argc, char *argv[]) {
if(argc < 2){
printf("Please enter a mode of operation.\n");
return 0;
}
else if(argc > 2){
printf("Too many arguments supplied. Please try again.\n");
return 0;
}
else if(strncmp(argv[1] == 'c') == 0)
calculate();
else if(strncmp(argv[1] == 'v') == 0)
verify();
return 0;
}
void calculate(){
}
void verify(){
}

Answer 1

You're not calling strncmp() correctly. It should be:
strncmp(argv[1], "c", 1)
It takes 3 arguments: two strings and a limit. 'c' is a char, not a string, string literals are put in double quotes.
You can also write simply:
if (argv[1][0] == 'c')
This just checks the first character, so the user can type y or yes and they'll match y. If you want to match the whole argument and require it be just a single character, you should use strcmp() rather than strncmp():
if (strcmp(argv[1], "c") == 0)

Answer 2

This
strncmp(argv[1] == 'c') == 0
just is incorrect syntax. For starters the function strncmp requires three arguments and it deals with strings while in the avove call of the function there is supplied only one argument of type int. because the wrong expression argv[1] == 'c' has type int and is equal to 0.
I can suggest the following solution
else if ( argc > 2 || strlen( argv[1] ) != 1 ){
printf("Too many arguments supplied. Please try again.\n");
return 0;
}
else if ( argv[1][0] == 'c' )
calculateCRC();
else if ( argv[1][0] == 'v' )
verifyCRC();