multiple inputs of multiple types

Im having trouble dealing with a program that takes multiple inputs (which can be a character, decimal, octal, hex) in any order. finding out if the input is a character or a digit.

When using argv it stores it as a string in a char array. I need to convert my input to a long or char depending on what it is. But I don't know the order, so i don't know what to expect. I trired using isalpha() but get Segmentation fault when executing.

If I knew what it was I could use strol(). Any ideas on how to deal with this??? Thank you.

Post the code you have along with any warnings and errors the compiler gives.

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>

void convert();

int main(int argc, char* argv[]){
        convert(argc,argv);
        return 0;
}

void convert(int size, char* element[]){

        unsigned i;
        for(i=1;i<size;i++){
                if(isalpha(element[i])){
                        char buffer;
                          strcpy(buffer,element[i]);
                        printf("%c  %d  %x\n", buffer,buffer,buffer);
                }
                else{
                        long buffer;
                        if(isxdigit(element[i])){
                                buffer = strtol(element[i],NULL,16);
                                printf("%c  %d  %x\n", buffer,buffer,buffer);
                        }
                        else{
                                strtol(element[i],NULL,10);
                                printf("%c  %d  %x\n", buffer,buffer,buffer);
                        }
                }
        }
}

---

during compile:

convert.c: In function `convert':
convert.c:18: warning: passing arg 1 of `strcpy' makes pointer from integer without a cast


Thanks.

You forgot to include string.h

You're having a hard time working with an array of pointers to char and you have signed mismatches. Then of course there is not including string.h. This compiles, but it won't work properly:

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>

void convert();

int main(int argc, char* argv[]){
  convert(argc,argv);
  return 0;
}

void convert(int size, char* element[]){

  int i;
  for(i=1;i<size;i++){
    if(isalpha(element[i][0])){
      char buffer[BUFSIZ];
      strcpy(buffer,element[i]);
      printf("%c  %d  %x\n", buffer,buffer,buffer);
    }
    else{
      long buffer;
      if(isxdigit(element[i][0])){
        buffer = strtol(element[i],NULL,16);
        printf("%c  %d  %x\n", buffer,buffer,buffer);
      }
      else{
        buffer = strtol(element[i],NULL,10);
        printf("%c  %d  %x\n", buffer,buffer,buffer);
      }
    }
  }
}

---

First glance things:
You are allocating only one char for buffer but are trying to use it as a string. That is the reason for

Quote:

---

convert.c:18: warning: passing arg 1 of `strcpy' makes pointer from integer without a cast

---

error

>1) Unless you are using a C99 compiler you can't declare variables after statements in C.
Variables can be declared only at the beginning of a block in C89, the OP's code is legal as this constraint is not violated anywhere in the posted code.

When testing I figured that out and removed it from my post. Thanks for the reminder though Prelude.

Thanks for the help, I think im on the right track now.