How to fix "incompatible integer to pointer conversion" in C

Question

I am trying to write C code to accept characters from a user one by one and then store them into a character array. If the user enters "1" I want the loop to terminate. The problem is that when I run the program, it just terminates and I get an error saying:

incompatible integer to pointer conversion passing 'char'
      to parameter of type 'const char *'; take the address with &
      [-Wint-conversion]
    while (strcmp(ch, "1") != 0)

How do I correct this? BTW, I'm using the WinGW-w64 compiler for Windows 10.

I have this code thus far:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MAX 100

int main(void)
{
    int i = 0;
    char ch, string[MAX];

    while (strcmp(ch, "1") != 0)
    {
        printf("Enter a character: ");
        scanf("%c", &ch);
        fflush(stdin);

        string[i] = ch;
        i++;
    }

    printf("%c", string[0]);
}

Please don't tag c++ in c questions, they're separate languages — Alan Birtles
– Alan Birtles, Commented Apr 21, 2021 at 7:46
As you have already been answered, strcmp is a function which compares 2 strings. The function expects to receive two const char *, but one of your arguments is a char. — Kfir Ventura
– Kfir Ventura, Commented Apr 21, 2021 at 8:01

Caleth · Accepted Answer · 2021-04-21 07:46:38Z

3

You have one character. strcmp requires a pointer to a sequence of characters that end with 0.

The easiest thing is to compare your one character to one character

while (ch != '1')

answered Apr 21, 2021 at 7:46

Caleth

66k2 gold badges53 silver badges101 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

chux Over a year ago

Another issue remains with char ch; ... while (ch != '1'), the initial value of ch is indeterminant.

Ketan Vadodariya · Accepted Answer · 2021-04-21 09:30:39Z

-1

Try this.

Solution.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MAX 100
    
    int main(void)
    {
        int i = 0, j = 0;
    
        char ch[MAX] = {0}, finalResult[MAX] = {0};
    
        //type cast with the strcmp argument
        //as per the function defination @strcmp will take "const char *", argument type
        while (strcmp((const char*)&ch, "1") != 0)
        {
            printf("Enter a character: ");
    
            scanf("%c", ch);   //as you are reading one 1 character so
    
            fflush(stdin);
    
            finalResult[j++] = ch[0];  //update user input character to the destination buffer
    
            i++; //increment for the putting '\0' to compare with "1"
    
            ch[i] = '\0';  //add null
    
        }
    
        printf("%c", finalResult[0]);
    }

answered Apr 21, 2021 at 9:30

Ketan Vadodariya

467 bronze badges

3 Comments

Lundin Over a year ago

Strictly speaking (const char*)&ch is wrong. If you take &ch where ch is an array, you get a pointer to the array of type char (*)[MAX]. This pointer is not compatible with const char*. So just drop the &ch and use ch.

Ketan Vadodariya Over a year ago

Let me clear why i took (const char *)&ch, first char ch[MAX] is local variable and function need a pointer argument so if you want to pass the local variable to any pointer function how you will pass it ? does typecast is necessary ? if not passing the location variable without casting does it work fine ? or garbage will come ?

Lundin Over a year ago

Arrays decay into a pointer to the first element whenever used in a normal expression like strcmp(ch, "1"). Pretty much the only exception to that array decay rule is when you use & on an array, which there is almost never a reason to do. Study arrays and pointers in your favourite C programming book.

Collectives™ on Stack Overflow

How to fix "incompatible integer to pointer conversion" in C

2 Answers 2

1 Comment

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related