C convert const char * to char - c

Question

I searched quite a while to find the answer, but I could only find a solution for C++ that didn't seem to work for C. I'm trying to convert argument of const char * to char to use in my switch statement. I tried various things like strdup(), but was unsuccessful.
#include <stdio.h>
int main(int argc, const char *argv[]) {
char argOne = argv[1];
char argTwo = argv[2];
switch(argOne) {
case '1234' :
printf("1234!\n" );
break;
default :
printf("Invalid\n" );
}
}
While compilation:
warning: incompatible pointer to integer conversion initializing 'char' with an expression of type 'const char *' [-Wint-conversion]
char argOne = argv[1];
^ ~~~~~~~
warning: overflow converting case value to switch condition type (825373492 to 52) [-Wswitch]
case '1234' :
^

Answer 1

In your code,
Point 1:
char argOne = argv[1];
is very wrong. You cannot put a const char * (pointer) to a char variable.
Point 2:
case '1234'
also wrong. Here, the '1234' does not denote a string. It is a multibyte charcater, which is most probably something you don't want. Again, even you change that to something like "1234", still it would be incorrect, as it will not give you the intended value, and strings cannot be used for case statement values.
Solution: Instead of using the switch case here, you can try using strcmp() to compare the incoming string and chose accordingly.
Note: The recommended signature of main() is int main(int argc, char *argv[])

Answer 2

You're getting mixed up between char (character) and char * (string). Also you can not use strings as switch/case labels. Here is a fixed version of your code:
#include <stdio.h>
int main(int argc, const char *argv[]) {
const char *argOne = argv[1];
const char *argTwo = argv[2];
if (strcmp(argOne, "1234) == 0)
{
printf("1234!\n");
}
else
{
printf("Invalid\n");
}
return 0;
}

Answer 3

First of all the standard declaration of main looks like
int main(int argc, char *argv[])
^^^^^^
That is the second parameter does not have qualifier const.
Secondly argv[1] has type char *. There is no any sense to compare it with a character literal similar to '1234'. As for string literal "1234" when it may not be used in the case label.
What you want is the following
#include <stdio.h>
#include <string.h>
int main(int argc, char *argv[]) {
if ( argc > 2 )
{
char *argOne = argv[1];
char *argTwo = argv[2];
if ( strcmp( argOne, "1234" ) == 0 )
{
puts( "1234!" );
}
else
{
puts( "Invalid" );
}
}
}

Answer 4

You can't (really) "convert" a pointer to char to a single char. You can however extract one single character from a string.
For example, to get the first character of the first argument to your program, you can do e.g.
char first_char_of_first_arg = 0;
if (argv > 1)
first_char_of_first_arg = argv[1][0];

Answer 5

the code contains several problems
compiling with all warnings enabled would allow the compiler to
display those problems.
for gcc, at a minimum, use: '-Wall -Wextra -pedantic'
Note: warnings need to be fixed, as the compiler knows the C language better than you or I.
#include <stdio.h>
int main(int argc, const char *argv[])
{
//<< parameter argc not used
// and it should be used to assure that argv[1] and argv[2] actually exist
char argOne = argv[1];
//<< argv[1] is a pointer to a character string.
// argOne is a single char
// it should be: 'char argOne = argv[0][0];'
char argTwo = argv[2];
//<< argv[2] is a pointer to a character string.
// argTwo is a single character
// it should be: 'char argTwo = argv[1][0];'
//<< argTwo is not used
switch(argOne)
//<< a character can be treated as an int, but due care needs to be exercised
{
case '1234' :
//<< a case statement can only look at an 'int',
// not a pointer to a (in this case unterminated) string
// and argOne is a single char, not a pointer to a unterminated string
printf("1234!\n" );
break;
default :
printf("Invalid\n" );
//<< for human readability, use consistent indentation
// and never use tabs for indenting
//<< every case should be terminated with 'break;'
// unless the case is to fall through to the next case
// I.E. just because a syntax 'can' be used does not mean it 'should' be used
}
}