How to remove all decimal places?

Can someone tell me why this code doesn't work... :confused:

Code:

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#include <stdio.h>

int main(void) {

        int a, i = 5.156546;

        a = trunc(i);

        printf("%d", a);

        return(0);
}

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And is there any other way to remove all numbers after a decimal place without using the trunc() function??

Well for a start trunc takes a double and returns a double, and you're playing with ints.

Secondly, i is an int (integer), so it can't store a value of 5.156546 (not an integer).

int a, i = 5.156546;
i = 5 because i is integer.
try
int a;
double d = 5.156546;
a = (int) d;
or print
printf("%.0f\n", d);

Quote:

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How to remove all decimal places?

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Cast to an int:

Code:

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/* prints "3" */
printf("%f", (int)3.14159265358979636);

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If you want to round the number, not truncuate it, use

Code:

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/* prints "3" */
printf("%d", (int)(3.14159265358979636 + .5));
/* prints "4" */
printf("%d", (int)(3.6 + .5));

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If you want to round the number to, say, 2 decimal places, use

Code:

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/* prints "3.14" */
printf("%f", (int)(3.14159265358979636 * 100 + .5) / 100.0);

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You don't really need the convoluted arithmetic:

Quote:

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Originally Posted by dwks

Cast to an int:
If you want to round the number, not truncuate it, use

Code:

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/* prints "3" */
printf("%d", (int)(3.14159265358979636 + .5));
/* prints "4" */
printf("%d", (int)(3.6 + .5));

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Code:

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/* prints "3" */
printf("%.0f", 3.14159265);
/* prints "4" */
printf("%.0f", 3.6);

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Quote:

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Originally Posted by dwks

If you want to round the number to, say, 2 decimal places, use

Code:

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/* prints "3.14" */
printf("%f", (int)(3.14159265358979636 * 100 + .5) / 100.0);

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Code:

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/* prints "3.14" */
printf("%.2f", 3.14159265358979);

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