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 http://stackoverflow.com/questions/2108192/what-are-the-valid-signatures-for-cs-main-function for details — , Dec 14 '13 at 18:56
On which operating system, with which compiler, and with which compiler options, are you compiling all this...? — Basile Starynkevitch, Dec 14 '13 at 19:36

Basile Starynkevitch · Answer 1 · 2013-12-14T19:37:29.757

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 '13 at 19:37

answered Dec 14 '13 at 18:53

Basile Starynkevitch

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Ok so how do I input and output float values then? :|| – SmOg3R Dec 14 '13 at 19:09
you don't have *float* values, unless you declare `float a[N+1], b[N+1], c[N+1];` – Basile Starynkevitch Dec 14 '13 at 19:33

haccks · Answer 2 · 2013-12-14T18:57:24.423

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 '13 at 18:57

answered Dec 14 '13 at 18:51

haccks

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score 2 · Accepted Answer · answered Dec 14 '13 at 18:53

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 '13 at 18:53

peterh

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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. – SmOg3R Dec 14 '13 at 19:04
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. – peterh Dec 14 '13 at 19:06
1

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 – SmOg3R Dec 14 '13 at 19:15

C float values conversion

3 Answers3