What is happening here in pow function?

Question

I have seen various answer here that depicts Strange behavior of pow function in C.
But I Have something different to ask here.

In the below code I have initialized int x = pow(10,2) and int y = pow(10,n) (int n = 2) .

In first case it when I print the result it shows 100 and in the other case it comes out to be 99 .

I know that pow returns double and it gets truncated on storing in int , but I want to ask why the output comes to be different.

CODE1

#include<stdio.h>
#include<math.h>
int main()
{
     int n = 2;
     int x;
     int y;
     x = pow(10,2);   //Printing Gives Output 100   
     y = pow(10,n);   //Printing Gives Output 99


     printf("%d %d" , x , y);

}

Output : 100 99

Why is the output coming out to be different. ?

My gcc version is 4.9.2

Update :

Code 2

int main()
    {
         int n = 2;
         int x;
         int y;
         x = pow(10,2);   //Printing Gives Output 100   
         y = pow(10,n);   //Printing Gives Output 99
         double k = pow(10,2);
         double l = pow(10,n);


         printf("%d %d\n" , x , y);
         printf("%f %f\n" , k , l);

    }

Output : 100 99
100.000000 100.000000

Update 2 Assembly Instructions FOR CODE1

Generated Assembly Instructions GCC 4.9.2 using gcc -S -masm=intel :

    .LC1:
    .ascii "%d %d\0"
    .text
    .globl  _main
    .def    _main;  .scl    2;  .type   32; .endef
_main:
    push    ebp
    mov ebp, esp
    and esp, -16
    sub esp, 48
    call    ___main
    mov DWORD PTR [esp+44], 2
    mov DWORD PTR [esp+40], 100      //Concerned Line
    fild    DWORD PTR [esp+44]
    fstp    QWORD PTR [esp+8]
    fld QWORD PTR LC0
    fstp    QWORD PTR [esp]
    call    _pow                    //Concerned Line
    fnstcw  WORD PTR [esp+30]
    movzx   eax, WORD PTR [esp+30]
    mov ah, 12
    mov WORD PTR [esp+28], ax
    fldcw   WORD PTR [esp+28]
    fistp   DWORD PTR [esp+36]
    fldcw   WORD PTR [esp+30]
    mov eax, DWORD PTR [esp+36]
    mov DWORD PTR [esp+8], eax
    mov eax, DWORD PTR [esp+40]
    mov DWORD PTR [esp+4], eax
    mov DWORD PTR [esp], OFFSET FLAT:LC1
    call    _printf
    leave
    ret
    .section .rdata,"dr"
    .align 8
LC0:
    .long   0
    .long   1076101120
    .ident  "GCC: (tdm-1) 4.9.2"
    .def    _pow;   .scl    2;  .type   32; .endef
    .def    _printf;    .scl    2;  .type   32; .endef

Answer 1

I know that pow returns double and it gets truncated on storing in int, but I want to ask why the output comes to be different.

You must first, if you haven't already, divest yourself of the idea that floating-point numbers are in any way sensible or predictable. double only approximates real numbers and almost anything you do with a double is likely to be an approximation to the actual result.

That said, as you have realized, pow(10, n) resulted in a value like 99.99999999999997 , which is an approximation accurate to 15 significant figures. And then you told it to truncate to the largest integer less than that, so it threw away most of those.

(Aside: there is rarely a good reason to convert a double to an int . Usually you should either format it for display with something like sprintf("%.0f", x) , which does rounding correctly, or use the floor function, which can handle floating-point numbers that may be out of the range of an int . If neither of those suit your purpose, like in currency or date calculations, possibly you should not be using floating point numbers at all.)

There are two weird things going on here. First, why is pow(10, n) inaccurate? 10, 2, and 100 are all precisely representable as double . The best answer I can offer is that the C standard library you are using has a bug. (The compiler and the standard library, which I assume are gcc and glibc, are developed on different release schedules and by different teams. If pow is returning inaccurate results, that is probably a bug in glibc, not gcc.)

In the comments on your question, amdn found a glibc bug to do with FP rounding that might be related and another Q&A that goes into more detail about why this happens and how it's not a violation of the C standard. chux's answer also addresses this. (C doesn't require implementation of IEEE 754 , but even if it did, pow isn't required to use correct rounding.) I will still call this a glibc bug, because it's an undesirable property.

(It's also conceivable, though unlikely, that your processor's FPU is wrong.)

Second, why is pow(10, n) different from pow(10, 2) ? This one is far easier. gcc optimizes away function calls for which the result can be calculated at compile time, so pow(10, 2) is almost certainly being optimized to 100.0 . If you look at the generated assembly code, you will find only one call to pow .

The GCC manual, section 6.59 describes which standard library functions may be treated in this way (follow the link for the full list):

The remaining functions are provided for optimization purposes.

With the exception of built-ins that have library equivalents such as the standard C library functions discussed below, or that expand to library calls, GCC built-in functions are always expanded inline and thus do not have corresponding entry points and their address cannot be obtained. Attempting to use them in an expression other than a function call results in a compile-time error.

[...]

The ISO C90 functions abort, abs, acos, asin, atan2, atan, calloc, ceil, cosh, cos, exit, exp, fabs, floor, fmod, fprintf, fputs, frexp, fscanf, isalnum, isalpha, iscntrl, isdigit, isgraph, islower, isprint, ispunct, isspace, isupper, isxdigit, tolower, toupper, labs, ldexp, log10, log, malloc, memchr, memcmp, memcpy, memset, modf, pow , printf, putchar, puts, scanf, sinh, sin, snprintf, sprintf, sqrt, sscanf, strcat, strchr, strcmp, strcpy, strcspn, strlen, strncat, strncmp, strncpy, strpbrk, strrchr, strspn, strstr, tanh, tan, vfprintf, vprintf and vsprintf are all recognized as built-in functions unless -fno-builtin is specified (or -fno-builtin-function is specified for an individual function).

So it would seem you can disable this behavior with -fno-builtin-pow .

Answer 2

Why is the output coming out to be different. ? (in the updated appended code)

We do not know the values are that different.

When comparing the textual out of int/double , be sure to print the double with sufficient precision to see if it is 100.000000 or just near 100.000000 or in hex to remove all doubt.

printf("%d %d\n" , x , y);
// printf("%f %f\n" , k , l);
// Is it the FP number just less than 100?
printf("%.17e %.17e\n" , k , l);  // maybe 9.99999999999999858e+01
printf("%a %a\n" , k , l);        // maybe 0x1.8ffffffffffff0000p+6

---

Why is the output coming out to be different. ? (in the original code)

C does not specify the accuracy of most <math.h> functions. The following are all compliant results.

// Higher quality functions return 100.0
pow(10,2) --> 100.0   
// Lower quality and/or faster one may return nearby results
pow(10,2) --> 100.0000000000000142...
pow(10,2) --> 99.9999999999999857...

Assigning a floating point (FP) number to an int simple drops the fraction regardless of how close the fraction is to 1.0

When converting FP to an integer, better to control the conversion and round to cope with minor computational differences.

// long int lround(double x);
long i = lround(pow(10.0,2.0));

Answer 3

You're not the first to find this. Here's a discussion form 2013: pow() cast to integer, unexpected result

I'm speculating that the assembly code produced by the tcc guys is causing the second value to be rounded down after calculating a result that is REALLY close to 100. Like mikijov said in that historic post, looks like the bug has been fixed.

Answer 4

As others have mentioned, Code 2 returns 99 due to floating point truncation. The reason why Code 1 returns a different and correct answer is because of a libc optimization.

When the power is a small positive integer, it is more efficient to perform the operation as repeated multiplication. The simpler path removes roundoff. Since this is inlined you don't see function calls being made.

Answer 5

你已经愚弄了它，认为输入是真实的，因此它给出了一个近似的答案，恰好略低于100，例如99.999999，然后被截断为99。