CODEWITHSUNDEEP
c
SmOg3R
SmOg3R

Reputation: 294

C float values conversion

int main(){

    int  N, i, j=0;

    float MA, MB, asum=0, bsum=0, y;

    printf("\number of pairs: "); scanf("%d", &N);

    int a[N+1], b[N+1], c[N+1];


    for(i=1; i<N+1; i++){

        printf("\na%d",i); printf("=");
        scanf("%f", &a[i]);


        printf("b%d",i); printf("=");
        scanf("%f", &b[i]);

        printf("\n aSUM= %.6f \n",asum);
        asum+=a[i];
        printf("\n aSUM= %.6f \n",asum);

    }

The idea of this code is simple. User inputs int or float values, then they get summed and outputted as a float value. However I'am getting astronomical values straight away. Fe. if it tries to make addition of 0 and 7, it outputs a value of 1088421888.000000. What the heck is going on?? :D

Upvotes: 0

Views: 107

Answers (3)

Basile Starynkevitch
Basile Starynkevitch

Reputation: 1

You should enable all warnings with a modern C compiler (like GCC 4.8).

 int a[N+1], b[N+1], c[N+1];
 //...later
     scanf("%f", &a[i]);

This cannot work: %f is for scanf(3) a control format requiring a float pointer, and &a[i] is a pointer to an int. (and GCC would have warned about that).

And you have other errors too. Please enable all warnings -e.g. compile with gcc -Wall

BTW, it is much better to put \n at the end of your printf format control string (not at the beginning!).

Upvotes: 5

haccks
haccks

Reputation: 106102

You are using wrong specifier for int. It will invoke undefined behavior. You will get anything. You are lucky that you are not getting the desired result!

Upvotes: 2

peterh
peterh

Reputation: 1

sscanf doesn't know about the type of their pointer parameters, so you read in floats in integer pointers. Thus, integer pointers were as float interpreted.

You need to scanf into a temporary float variable, and then convert this to integers. So:

float  theFloat;
sscanf("%f", &theFloat);
N[a]=theFloat;

Upvotes: 2

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