Why does gdb evaluate sqrt(3) to 0?

Question

The square root of 3, as estimated by Wolfram Alpha:

1.7320508075688772935274463415058723669428052538103806280558...

When I do sqrt(3) in C, it evaluates to 0. Why?

EDIT4: here's how you can reproduce this issue in GDB. Create test.c as follows:

#include <stdio.h>                                                                                                                                                                      
#include <math.h>

int main()
{
  printf("sqrt(3): %f\n", sqrt(3));
  return 0;
}

Compile:

gcc -O0 -g -Wall -pedantic -ansi -lm -o test test.c

Run debugger:

gdb test

Enter this at console:

(gdb) break test.c:6
Breakpoint 1 at 0x400578: file test.c, line 6.
(gdb) r
Starting program: /home/pdedecker/Desktop/test   
Breakpoint 1, main () at test.c:6
6         printf("sqrt(3): %f\n", sqrt(3));
(gdb) print sqrt(3)
$1 = 0
(gdb) s
sqrt(3): 1.732051

My GDB version is GNU gdb (GDB) SUSE (7.1-3.12).

Answer 1

The problem is not the missing function declaration (which isn't missing, since you did include <math.h>).

The problem is missing debug info for the sqrt you are actually using. Without that debug info, GDB has no clue what parameter type to pass to sqrt(), and what it returns.

You can get the required debug info on many Linux distributions by installing libc-debuginfo package. Here is what I see on such a system:

gdb -q ./a.out
Reading symbols from /tmp/a.out...done.
(gdb) b main
Breakpoint 1 at 0x400558: file t.c, line 6.
(gdb) r

Breakpoint 1, main () at t.c:6
6     printf("sqrt(3): %f\n", sqrt(3));
(gdb) p sqrt
$1 = {<text variable, no debug info>} 0x7ffff7b7fb50 <__sqrt>

Note: "no debug info"

(gdb) p sqrt(3)
$2 = 0
(gdb) p sqrt(3.0)
$3 = 0

Note: matches your behavior. What sqrt functions do have debug info?

(gdb) info func sqrt
All functions matching regular expression "sqrt":

File ../sysdeps/x86_64/fpu/e_sqrt.c:
double __ieee754_sqrt(double);

File s_csqrt.c:
complex double __csqrt(complex double);

File ../sysdeps/x86_64/fpu/e_sqrtf.c:
float __ieee754_sqrtf(float);

File w_sqrtf.c:
float __sqrtf(float);

File s_csqrtf.c:
complex float __csqrtf(complex float);

File ../sysdeps/i386/fpu/e_sqrtl.c:
long double __ieee754_sqrtl(long double);

File w_sqrtl.c:
long double __sqrtl(long double);

File s_csqrtl.c:
complex long double __csqrtl(complex long double);

File ../sysdeps/ieee754/dbl-64/mpsqrt.c:
void __mpsqrt(mp_no *, mp_no *, int);

File w_sqrt.c:
double __sqrt(double);

(gdb) p __sqrt
$4 = {double (double)} 0x7ffff7b7fb50 <__sqrt>

Note: __sqrt is at the same address as sqrt, but GDB knows its type!

(gdb) p __sqrt(3)
$5 = 1.7320508075688772
(gdb) p __sqrt(3.0)
$6 = 1.7320508075688772

One can reasonably argue this is a bug in GDB. Feel free to create one in GDB bugzilla.

Answer 2

I predict that you didn't do #include <math.h>

Without a function declaration, C will default the return value of a function to int. A floating point number might come back as 0 depending on the size of your int. C will also not know how to convert the function argument. It will default to passing the argument as whatever type it happens to be. If you pass an integer to sqrt() it will not be converted to a double, but the sqrt() function will interpret the bit pattern as double.

Answer 3

To call a function without debug info, you must explicitly tell gdb the type for the return and arguments, using a function pointer cast. So, for your example:

(gdb) print ((double (*) (double)) sqrt) (3)
$1 = 1.7320508075688772