A Char variable in C changing value

Question

#include<stdlib.h>
#include<stdio.h>
int main()
{
    char v1=0;
    char v2=0;
    printf("Enter the number of elements of first vectors : ");
    scanf(" %d",&v1);
    printf("v1=%d\n",v1);
    printf("Enter the number of elements of second vectors : ");
    scanf(" %d",&v2);
    printf("v2=%d\n",v2);
    printf("v1=%d\n",v1);
    printf("v2=%d\n",v2);
    return 0;
}

If we assumed v1=50 and v2=300 Why does v1 value changes in the second print

If your compiler isn't yelling at you, turn up the warning levels. — Mat, Feb 22 '22 at 21:15
You can print it with `%d` because of default argument promotions, but that doesn't work with pointers. — Barmar, Feb 22 '22 at 21:17
Re “A Char variable in C changing value”: Changing values is what variables do. It is their primary purpose to vary. You should write a more descriptive and specific title. — Eric Postpischil, Feb 22 '22 at 21:27
You may want to read this: [Why should I always enable compiler warnings?](https://stackoverflow.com/q/57842756/12149471) — Andreas Wenzel, Feb 22 '22 at 21:54

score 1 · Answer 1 · answered Feb 22 '22 at 21:18

1

here is what my compiler says with warning turned up max

warning C4477: 'scanf' : format string '%d' requires an argument of type 'int *', but variadic argument 1 has type 'char *'

you need

scanf(" %c",&v1);

answered Feb 22 '22 at 21:18

pm100

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It works find in my compiler , I get the first printf output normally but the second for v1 comes out as 1 – Optix Feb 22 '22 at 21:25
@Optix OK well you have a poor compiler, you might need to ramp up the warning level. I also provided the fix for you, did you try that – pm100 Feb 22 '22 at 21:32
@Optix google "c undefined bahaviour" – Jabberwocky Feb 22 '22 at 21:47

A Char variable in C changing value

1 Answers1