C将const char *转换为char

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 eg

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
    }
}