Code coverage

rajsun · 06-29-2005, 03:57 AM

Hi all,
I m useing 'gcov' for first time.
The code is :
========================================================================
#include<stdio.h>

void Disply1(int sum)
{
printf("The sum %d less than 100\n",sum);
}

void Disply2(int sum)
{
printf("The sum %d more than 100\n",sum);
}

int main()
{
int i=0;
int sum = 0;

for(; i<10; )
{
sum+=5;
i++;
}

if (sum <= 100)
{
Disply1(sum);
}
else
{
Disply2(sum);
}
return 0;
}
==============================================================================
I compiled the code :
gcc -fprofile-arcs -ftest-coverage cfile.c

then I executed the code :
./a.out

then i typed the following commend :
gcov -b cfile.c

And I got the following code :
==============================================================================
#include<stdio.h>

void Disply1(int sum)
1 {
1 printf("The sum %d less than 100\n",sum);
call 0 returns = 100%
}

void Disply2(int sum)
###### {
###### printf("The sum %d more than 100\n",sum);
call 0 never executed
}

int main()
1 {
1 int i=0;
1 int sum = 0;

21 for(; i<10; )
branch 0 taken = 91%
branch 1 taken = 100%
branch 2 taken = 100%
{
10 sum+=5;
10 i++;
}

1 if (sum <= 100)
branch 0 taken = 0%
{
1 Disply1(sum);
call 0 returns = 100%
branch 1 taken = 100%
}
else
{
###### Disply2(sum);
call 0 never executed
}
1 return 0;
}
==============================================================================

Problem :
I cant understand

Why it is giving :
-----------------------------
21 for(; i<10; )
branch 0 taken = 91%
branch 1 taken = 100%
branch 2 taken = 100%
{
10 sum+=5;
10 i++;
}
-------------------------------------

1-> what is the meaning of branch 0, branch 1 and branch 2.
2-> why for statement has been executed 21 time?

Plz hepl me I could not understand the cause.

Tahnking U all,
With Regard
Rajsun

Mara · 06-29-2005, 03:14 PM

gcov manual says

Code:

       For each basic block, a line is printed after the last line of the
       basic block describing the branch or call that ends the basic block.
       There can be multiple branches and calls listed for a single source
       line if there are multiple basic blocks that end on that line.  In this
       case, the branches and calls are each given a number.  There is no sim-
       ple way to map these branches and calls back to source constructs.  In
       general, though, the lowest numbered branch or call will correspond to
       the leftmost construct on the source line.

My guess (and it's a good guess) is that in line
for(; i<10; )
the three branches are (empty) then i < 10; and then (empty).

21 runs - don't know why (if you sum up execution of the three branches above, the result is 21, but it's probably not the case).

vladmihaisima · 06-29-2005, 05:28 PM

You can use:

gcc -g cfile.c
objdump -S a.out > cfile.asm

And look for the line with the for loop. My disassembly looks like:

for(; i < 10

{
804837c: 83 7d fc 09 cmpl $0x9,0xfffffffc(%ebp)
8048380: 7e 02 jle 8048384 <main+0x26>
8048382: eb 0d jmp 8048391 <main+0x33>
sum += 5;
8048384: 8d 45 f8 lea 0xfffffff8(%ebp),%eax
8048387: 83 00 05 addl $0x5,(%eax)
i++;
804838a: 8d 45 fc lea 0xfffffffc(%ebp),%eax
804838d: ff 00 incl (%eax)
804838f: eb eb jmp 804837c <main+0x1e>
}
if(sum <= 100) {
8048391: 83 7d f8 64 cmpl $0x64,0xfffffff8(%ebp)

So the first branch is the jle (jump less equal) which executes only 10 time from 11 executions (0.909% aprox 0.91%)
I assume the the second branch is the fall branch (jle not taken so flow goes to the next instruction).
The third branch must be the jmp.

The execution count could be computed on basic blocks and in the for instruction there are 2 basic blocks (cmp and jle one, and jmp another) So 11 for the first and 10 for the second.

Hope it helps.