C float values conversion

Question

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

Remember, if this is C and not C++ your definition of main() must be int main(void) or int main(int argc, char* argv[]). see stackoverflow.com/questions/2108192/… for details — user2363448
– user2363448, Commented Dec 14, 2013 at 18:56
On which operating system, with which compiler, and with which compiler options, are you compiling all this...? — Basile Starynkevitch
– Basile Starynkevitch, Commented Dec 14, 2013 at 19:36

Basile Starynkevitch · Accepted Answer · 2013-12-14 19:37:29Z

5

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!).

edited Dec 14, 2013 at 19:37

answered Dec 14, 2013 at 18:53

Basile Starynkevitch

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2 Comments

SmOg3R Over a year ago

Ok so how do I input and output float values then? :||

Basile Starynkevitch Over a year ago

you don't have float values, unless you declare float a[N+1], b[N+1], c[N+1];

peterh · Accepted Answer · 2013-12-14 18:53:07Z

2

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;

answered Dec 14, 2013 at 18:53

peterh

1

3 Comments

SmOg3R Over a year ago

O so now if I input 5.6 it gets converted to 5, therefore 0+5.6 becomes 5... I might as well just use ints and get the same result.. I need to input and output values with commas.

peterh Over a year ago

No! You have right! You need to handle the main array as a float array! Your current code rounds the floats during the integer conversion.

SmOg3R Over a year ago

Thanks! I don't get it why people tend to answer in such a complex way lol. "change {int a[N+1]} to {float a[n+1]}" is all that needed. Thanks again

haccks · Accepted Answer · 2013-12-14 18:57:24Z

2

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!

edited Dec 14, 2013 at 18:57

answered Dec 14, 2013 at 18:51

haccks

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