https://gcc.gnu.org/bugzilla/show_bug.cgi?id=101191

            Bug ID: 101191
           Summary: [GCOV] Wrong coverage with "for(;;)" statement
           Product: gcc
           Version: 10.2.0
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: gcov-profile
          Assignee: unassigned at gcc dot gnu.org
          Reporter: njuwy at smail dot nju.edu.cn
                CC: marxin at gcc dot gnu.org
  Target Milestone: ---

$ gcc -v
Using built-in specs.
COLLECT_GCC=gcc
COLLECT_LTO_WRAPPER=/usr/local/libexec/gcc/x86_64-pc-linux-gnu/10.2.0/lto-wrapper
Target: x86_64-pc-linux-gnu
Configured with: ../configure -enable-checking=release -enable-languages=c,c++
-disable-multilib
Thread model: posix
Supported LTO compression algorithms: zlib
gcc version 10.2.0 (GCC) 

$ cat test.c
#include<stdio.h>
#include<stdlib.h>
int x[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};

int main() {
  int niterations = 0, i;

  for (;;) {
    int i, mi, max;
    max = 0;
    for (i = 0; i < 10; i++) {
      if (x[i] > max) {
        max = x[i];
        mi = i;
      }
    }
    if (max == 0)
      break;
    x[mi] = 0;
    niterations++;
    if (niterations > 10)
      abort();
  }

  exit(0);
}

$ gcc -O0 --coverage test.c;./a.out;gcov test;cat test.c.gcov
libgcov profiling error:/home/wangyang/coverage/test/test.gcda:overwriting an
existing profile data with a different timestamp
File 'test.c'
Lines executed:93.33% of 15
Creating 'test.c.gcov'

        -:    0:Source:test.c
        -:    0:Graph:test.gcno
        -:    0:Data:test.gcda
        -:    0:Runs:1
        -:    1:#include<stdio.h>
        -:    2:#include<stdlib.h>
        -:    3:int x[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
        -:    4:
        1:    5:int main() {
        1:    6:  int niterations = 0, i;
        -:    7:
        9:    8:  for (;;) {
        -:    9:    int i, mi, max;
       10:   10:    max = 0;
      110:   11:    for (i = 0; i < 10; i++) {
      100:   12:      if (x[i] > max) {
       45:   13:        max = x[i];
       45:   14:        mi = i;
        -:   15:      }
        -:   16:    }
       10:   17:    if (max == 0)
        1:   18:      break;
        9:   19:    x[mi] = 0;
        9:   20:    niterations++;
        9:   21:    if (niterations > 10)
    #####:   22:      abort();
        -:   23:  }
        -:   24:
        1:   25:  exit(0);
        -:   26:}

Line 8 should be executed 10 times

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