1

I'm testing conversion specifiers both for printf() and scanf() function. I hadn't problem with %a format specifier for floating point hexadecimal number used in printf() function. It displayed what I expected. But I encountered problem when I wanted to input hexadecimal float using

scanf("%a", &var);

I'm using the following code:

#include <stdio.h>

int main()
{
    float var;

    printf("Enter variable:\n");
    scanf("%a", &var);
    printf("var = %f\n", var);
    printf("var = %e\n", var);
    printf("var = %a\n", var);

    return 0;
}

When I enter a sample value, e.g. 0x1.205b0cp-11 that is 5.5e-4 it gives me that result:

Enter variable: 0x1.205bc0p-11

var = 0.000000 var = 0.000000e+000 var = 0x0.000000p+0

Whatever number I enter it always gives me 0. What's the reason or solution for that problem. I have no idea what I'm making wrong.

Karol Chojecki
  • 11
  • 2
  • 2
    It works for me. Always check the value returned by `scanf`; it tells you how many items it successfully read. – Keith Thompson Feb 09 '23 at 21:38
  • Works for me on my machine, too. – Steve Summit Feb 09 '23 at 21:51
  • 1
    Might be worth testing with `scanf("%f", &var);` to see if that works, to see if maybe the problem is that your system's `scanf` doesn't properly support `%a`. What is your system and compiler, anyway? – Steve Summit Feb 09 '23 at 21:54
  • @Steve Summit I'm using `gcc (x86_64-posix-seh-rev0, Built by MinGW-W64 project) 8.1.0` running on `Windows 10 Version 21H2`. According to what you pointed, `scanf("%f", &var)` gives me proper result if I input decimal floating point. – Karol Chojecki Feb 09 '23 at 23:31
  • @KarolChojecki Under Windows, I believe that gcc/mingw use some of the Microsoft C libraries, and those libraries are known to have some significant deficiencies. I hadn't heard of total failure to parse `%a`, but I suppose it's possible. – Steve Summit Feb 09 '23 at 23:40
  • The input to `%a` is supposed to begin with `0x`, but it might be worth trying just `1.205b0cp-11` instead. (Although, when I tried that, I got `1.205` instead of `5.5e-4`.) – Steve Summit Feb 09 '23 at 23:42
  • Please see [this question](https://stackoverflow.com/questions/23900921) and [Keith Thompson's answer there](https://stackoverflow.com/questions/23900921#23901476). – Steve Summit Feb 09 '23 at 23:59

2 Answers2

1

That code works on my system(a), producing the expected result:

pax> ./prog
Enter variable: 0x1.205bc0p-11
var = 0.000550
var = 5.500000e-04
var = 0x1.205bcp-11

Keep in mind that hexadecimal floating point constants (and the %a conversion specifier) were introduced in C99, so you should make sure you're using a compiler that handles that. Most modern ones will, but some people may be locked in to a certain level if the vendor isn't keen on keeping up to date(b).

Keep in mind that the scanf family, conversion specifiers a, e, g, and f (and their upper-case variants) are all the same - they each handle every format that strtod handles. The (likely) reason that they all exist is to mirror the printf conversion specifiers where it makes a difference to how the value is output.

It's possibly worth testing with a variety of inputs less advanced than hex floating point, and work your way up to there. For example, the sequence { 0, 3, 3.1, 3.14159. 9e2, 6e-2, ... }. That would at least let you know where the input is failing.

Also, it's a good idea to test the conversion to ensure it works (this is true of all calls where failure can adversely affect future program behaviour). In other words, something like this in place of your scanf statement:

if (scanf("%a", &var) != 1) {
    fprintf(stderr, "Something went horribly wrong!\n");
    return 1;
}

(a) My system, by the way, is gcc 9.4.0 (from gcc --version) running on Ubuntu 20.04.2 (from cat /etc/lsb_release). It's often a good idea to include that sort of information in questions, it can help us figure out issues more quickly.

(b) Possibly a good reason to consider changing vendors though, of course, subject to normal cost/benefit analysis.

paxdiablo
  • 854,327
  • 234
  • 1,573
  • 1,953
  • Sorry, this is my first question. I'm using `gcc (x86_64-posix-seh-rev0, Built by MinGW-W64 project) 8.1.0` running on `Windows 10 Version 21H2`. – Karol Chojecki Feb 09 '23 at 23:31
0

Try computing somehow like with printf("var = %e\n", var/7.0 + var *var);. Sometimes simple code without doing any math does not link in the math library needed by printf().

chux - Reinstate Monica
  • 143,097
  • 13
  • 135
  • 256