Casting float to (int) in C - declaring cast (int) seems to have no effect

Question

C program takes float value as command line argument, so need to format from string to float and then to integer. Using round() from math.h, and then want to cast to int.

Casting by declaring (int) before value, but value type does not change.

Code below:

double *f = (double *) malloc(sizeof(double));
*f = (int)roundf(atof(argv[1]));
printf("Testing f: %d", *f);

make gives this error message:

format specifies type 'int' but the argument has type 'double'
      [-Werror,-Wformat]
    printf("Testing f: %d", *f);

What type should be changing? Also why are you casting at all? — UnholySheep, Apr 28 '17 at 19:15
You can't change a type at runtime - C is a statically typed language — UnholySheep, Apr 28 '17 at 19:16
You can specify rounding direction (up/down) with `fenv`. The typical round `.5` up to the next integer value can be expressed (when rounding `fv` as the float value to integer `v`) as `v = (int)(fv > 0 ? fv + .5 : fv - .5);` to handle both positive and negative cases. — David C. Rankin, Apr 28 '17 at 20:45
@DavidC.Rankin `v = (int)(fv > 0 ? fv + .5 : fv - .5);` is a problem (wrong answer) for a number of values of `fv` when the `fv + .5` incurs a rounding before the cast `(int)`. C has functions that do the conversion better: `round(), rint(), lround(), nearby()`. — chux - Reinstate Monica, Apr 28 '17 at 21:15
Yes, I should qualify that by saying the purpose of the math library functions are to provide protections against a number of cases that do not work with a simple ternary, namely `INF`, `NAN`, etc. Good catch. I guess `typical` was not apparent enough to explain the comment was by way of an example. — David C. Rankin, Apr 28 '17 at 21:19

score 3 · Answer 1 · edited Apr 28 '17 at 19:22

3

You're putting int into double. The f variable should be of type int.

int f;
f = (int)roundf(atof(argv[1]));

edited Apr 28 '17 at 19:22

David Ranieri

39,972
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answered Apr 28 '17 at 19:16

martin49

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The cast is redundant. – pmg Apr 28 '17 at 20:19
2

@pmg Pedantic compilers warning oblige an `(int)` cast to quiet warnings such as "conversion to 'int' from 'float' may alter its value [-Wfloat-conversion]" – chux - Reinstate Monica Apr 28 '17 at 20:38

score 1 · Answer 2 · answered Apr 28 '17 at 20:35

C has a direct way to round and convert a float to an integer type

long int lround(double x);

The lround and llround functions round their argument to the nearest integer value, rounding halfway cases away from zero, regardless of the current rounding direction. ... C11 7.12.9.7 2

#include <math.h>
#include <stdlib.h>

long i = lround(atof(argv[1]));
// or 
float f = atof(argv[1]);
long i = lroundf(f);

// Use %.0f to print a FP with no factional digits.
printf("%.0f\n", f);

// Use %ld to print a `long`
printf("%ld\n", i);

score 1 · Answer 3 · answered Apr 28 '17 at 21:29

The error is in the %d format specifier in the printf function:

printf("Testing f: %d", *f); /* %d expects an integer but *f is a double */

It doesn't matter that *f contains a rounded number, it is still a double as far as printf is concerned.

Try this:

printf("testing f: %d", (int)*f);

N.B. why you are going to the trouble of using malloc to allocate a single double? If you need to pass just one double to some other program, you could just have:

double f;
... stuff ...
foo(&f);

Casting float to (int) in C - declaring cast (int) seems to have no effect

3 Answers3