CODEWITHSUNDEEP
c
Akhil Kumar
Akhil Kumar

Reputation: 11

error: comparision between pointer and integer

#include<stdio.h>
#include<string.h>
int main(int argc,char *argv[])
{
char c;
char a[19][100];
int i=0;
int j=0;

while((c=fgetc(stdin))!=EOF)
{
  if(c!=" ")
  {
    a[i][j]=c;
    j++;
   }
   if(c==" "||c=='\n')
{
   a[i][j]='\0';
   i++;
   j=0;
    }
 }
 for(j=0;j<i;j++)
printf("%s \n",a[j]);
}

the error that i get is

   15.c:12:5: warning: comparison between pointer and integer[enabled       by   default]

 if(c!=" ")
     ^

15.c:17:5: warning: comparison between pointer and integer [enabled by   default]

if(c==" "||c=='\n')

Upvotes: 0

Views: 666

Answers (2)

RadijatoR
RadijatoR

Reputation: 113

You compare char with char *. You can not do it, and set c to be signed char because it need to store EOF. It depends on compiler does char will signed or unsigned, so you can't know. It need to be like this: 

if(c!= ' ')
  {
    a[i][j]=c;
    j++;
   }

Upvotes: -1

alk
alk

Reputation: 70931

c is defined to be a char and " " is a "string" literal, which in C is implemtned as an array of chars, which in turn decays to a pointer to it's 1st element, that is a char *.

To fix this compare c to a char.

if (c != ' ')  /* Use ' to code a char literal, use " to code a string literal. "/

Also as you assign the result of fgetc() to c and test c against EOF (which typically is -1) make c an int, that is make it the type fgetc() returns.

Upvotes: 2

Related Questions