My program runs but I get some warnings that tell me the format %d can't go there

Question

I have to make a program which asks me to enter different values and then it has to print them on the screen. Aftar that, It asks me, "which value of the vector you want to change?" and I have to enter the number of the value and than the value itsself.

When I run the program I get several warnings like this:

"warning: format '%d' expects argument of type 'int', but argument 6 has type 'int *' [-Wformat=]"

I don't know how to fix this, I tried using %p but it doesn't help. Any ideas?

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

#define MAX 5

int main()
{
    int intero = 0;
    int *p_int;

    int vettore[MAX] = {0};
    int *p_vet;
    int num = 0;
    int val = 0;

    char carattere = 0;
    char *p_car;

    p_int = &intero;
    p_car = &carattere;
    p_vet = vettore;

    printf("Enter an integer: ");
    scanf ("%d", &intero);
    fflush (stdin);

    printf ("\nEnter a character: ");
    scanf("%c", &carattere);
    fflush (stdin);

    for (int i=0; i<MAX; i++)
    {
        printf("\nEnter integer n^%d of the vector: ", i);
        scanf("%d", &vettore[i]);
        fflush (stdin);
    }

    printf("\n\n\n");

    printf("Type\t\tDirect Value\t\tValue by Pointer\tDirect Position\tPosition by Pointer\n");
    printf("__________________________________________________________________________________________________________________\n\n");
    printf("Integer:\t\t%d\t\t\t%d\t\t\t\t%d\t\t\t\t%d\n\n", intero, *p_int, &intero, p_int);
    printf("Character:\t%c\t\t\t%c\t\t\t\t%d\t\t\t\t%d\n\n", carattere, *p_car, &carattere, p_car);

    for (int i=0; i<MAX; i++)
    {
        printf("Integer n^%d:\t%d\t\t\t%d\t\t\t\t%d\t\t\t\t%d\n\n", i, vettore[i], *p_vet, &vettore[i], p_vet);
        p_vet++;
    }

    p_vet = vettore;

    printf("\nEnter which value of the vector you want to change: ");
    scanf("%d", &num);
    fflush(stdin);

    printf("\nEnter now the value you want: ");
    scanf("%d", &val);
    fflush(stdin);

    *(p_vet + num) = val;
    printf("\n\n");

    for (int i=0; i<MAX; i++)
    {
        printf("Integer n^%d:\t%d\t\t\t%d\t\t\t\t%d\t\t\t\t%d\n\n", i, vettore[i], *p_vet, &vettore[i], p_vet);
        p_vet++;
    }

    return 0;
}

Aside: `fflush(stdin)` isn't allowed, and isn't necessary except to remove the newline input before `scanf("%c", &carattere);`. Please add a space so it becomes `scanf(" %c", &carattere);` and that will filter the newline. — Weather Vane, Nov 13 '20 at 09:22
Show your attempt with `%p`. `%p` must be used with pointers. — Jabberwocky, Nov 13 '20 at 09:23
`printf("Integer:\t\t%d\t\t\t%d\t\t\t\t%p\t\t\t\t%p\n\n", intero, *p_int, &intero, p_int);` works for me. — Jabberwocky, Nov 13 '20 at 09:24
It's vaguely possible that a cast before a pointer passed for `%p` is needed: `... (void*)&vettore[i], (void*)pvet)`. Also there are three `printf()`s that need changing. — Weather Vane, Nov 13 '20 at 09:32

Nastor · Accepted Answer · 2020-11-13T09:36:10.543

By using %p format specifier, you're telling printf that the value you're trying to print is actually a memory address, not the value the pointer points to. Whenever you use a pointer without dereferencing it with * operand, you're just printing the memory offset it points to, which isn't formatted as integer. It seems to me that sometimes you just do that, like here: printf("Integer:\t\t%d\t\t\t%d\t\t\t\t%d\t\t\t\t%d\n\n", intero, *p_int, &intero, p_int);

Since p_int is a pointer and you're using %d as format specifier, it throws you that kind of error. Dereferencing it with *p_int is the right thing to do, I think, unless your intention is to actually print the memory offset, in that case you should use %p and cast the pointer to (void*).

Figurati, I'm italian too by the way ;) – Nastor Nov 13 '20 at 09:35 — Nastor, Nov 13 '20 at 09:35

score 1 · Answer 2 · answered Nov 13 '20 at 09:33

The exact compiler output is:

warning: format '%d' expects argument of type 'int', but argument 4 has type 'int *' [-Wformat=]
42 |     printf("Integer:\t\t%d\t\t\t%d\t\t\t\t%d\t\t\t\t%d\n\n", intero, *p_int, &intero, p_int);
   |                                           ~^                                 ~~~~~~~
   |                                            |                                 |
   |                                            int                               int *

Notice those "ASCII arrows" pointing out first the location of the format specifier and then the location of the parameter that is of the wrong type. In plain English it says BUG HERE and BUG HERE. Hmm, where could the bug be? Possibly at those locations!

In this case it turns out that %d expects int but &intero is int*. Just like the compiler spelled out to us.

I tried using %p but it doesn't help

It helps tremendously, since it makes the bugs go away. I did this on your very code and got rid of all the warnings. All I did was to replace d with p at the exact locations the compiler pointed out to me. I literally did not need to activate my brain at all to find and fix the bugs, I just blindly followed the compiler's advise and fixed the exact locations in the format strings pointed out to me.

Yes I was using %p in the wrong position. I'm a beginner, thanks for the help! — ilMichigan, Nov 13 '20 at 09:37

score 0 · Answer 3 · answered Nov 13 '20 at 09:25

0

In the lines containing "scanf" remove the "&". This causes the code to pass a reference to the value instead of the actual value. See Meaning of "referencing" and "dereferencing" in C for an explanation of referencing

answered Nov 13 '20 at 09:25

DevL

11
3

3

I don't want to downvote this because you don't have many votes. The `&` is used correctly (none of the formats is `%s` and none of the arguments is already a pointer). – Weather Vane Nov 13 '20 at 09:26

My program runs but I get some warnings that tell me the format %d can't go there

3 Answers3