Why %.2f like syntax not working

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i want to use syntax like %.2f which means it has to print upto two decimal places but its not taking like that instead of that it is not even reading as float value why?

What I have tried:

printf("ans=%.2f\n",a/b)

Posted 21-Feb-17 8:04am

Member 13015247

Updated 21-Feb-17 8:44am

v2

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Richard MacCutchan 21-Feb-17 14:26pm

What types are a and b?

Member 13015247 21-Feb-17 14:41pm

a and b are declared as floats initially and ya this problem is occuring in cs50ide why? in codeblock its running good

Richard MacCutchan 21-Feb-17 14:43pm

So your actual issue is that cs50ide does not work. Pity you did not state that in the first place.

2 solutions

Solution 2

Make sure a/b is a float.
If a and b are integers, / result in an integer.

Posted 21-Feb-17 8:44am

Patrice T

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OriginalGriff · Accepted Answer · 2017-02-21T08:29:00

Solution 1

Probably because your code is using integer values for a and b. Try declaring them as float, not int:

#include <stdio.h>

int main()
{
    float a = 24;
    float b = 7;
    printf("Hello, World!\n");
    printf("ans = %.2f\n", a / b);
    printf("ans = %2.2f\n", a / b);
    return 0;
}

Posted 21-Feb-17 8:29am

OriginalGriff

Comments

Richard MacCutchan 21-Feb-17 14:43pm

See comments above.

OriginalGriff 21-Feb-17 14:57pm

See his other question...:laugh:

Member 13015247 21-Feb-17 15:03pm

for which ques r u talking?

Richard MacCutchan 21-Feb-17 15:41pm

All of them.

Member 13015247 21-Feb-17 15:05pm

if u can solve the problem then solve it otherwise laugh somewhere else this is not the place to laugh