I'm trying to trying to print the output but my code keep getting c compile error: ISO C forbids comparison between pointer and integer, The error is in else if :

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

int main (){
char BT [5][100];



for (int i = 1; i<=2; i  ){
        printf("\nInsert book title:");scanf("%[^\n]", &BT[i]); getchar ();

}

    printf("\nOur Collections :\n"); 
    
    for (int i = 1; i<= 2 ; i  ){   
    for (int i = 0; i< strlen(BT[i]); i  ){
            int k;
        if ( i == 0 && BT[i][k] != ' ') {
            printf("Shelf code : %c\n", BT[i][k]);
        }
        else if ( i > 0 && BT[i - 1] == ' ') {
            printf("Shelf code : %c\n", BT[i][k]);
    }

}
}
return 0;
}

Thanks in advance!

CodePudding user response：

1. You compile C program using C compiler which is wrong.
2. BT[i - 1] is an char array acting as lvalue (pointer) not character. You compare it to an integer ' '. Even in C it will raise a warning and it is definitely not what you want.

CodePudding user response：

There are many mistakes.

For example indices in C for arrays start from 0 but you are using for loops where indices start from 1 like

for (int i = 1; i<=2; i  ){

and below in the code you are trying to access BT[0].

Or the function scanf for reading a string expects an argument of the type char * but you are passing an argument of the type char ( * )[100].

scanf("%[^\n]", &BT[i]);

At least you should write

scanf("