Program Crashes - Segmentation Fault

Question

My Code

int i,v,o,p,j;
v=0;
o=0;
char x[12]={'#','v','v','o','#','o','v','#','v','o','o','#'};
void vukovi (char a[])
{
    for (i=0;i<12;i++){
        if (x[i]=='#'){
            for (j=i+1;x[j]!='#';j++){
                if (x[j]=='v'){
                    v=v+1;
                }
                if (x[j]=='o')
                    o=o+1;
            }

            if (v>=o){
                for (j=i+1;x[j]!='#';j++){
                    if (x[j]=='o'){
                        x[j]='.';
                    }
                }
            }

        }
        v=0;
        o=0;
    }

for (i=0;i<12;i++){
    printf("%c",x[i]);
    }
}
vukovi(x);
return 0;}

Sometimes it prints well.
Sometimes error

.exe stopped working

I have no idea why.
Idea of program is : everywhere between two # where is v>o , o must be replaced with .
Like I said when it prints, it prints well.
Otherwise error. Help ?
( I Have left out include and main function ).

[Title translation](https://en.wikipedia.org/wiki/Undefined_behavior) — LPs, Feb 09 '17 at 13:12

Jean-François Fabre · Answer 1 · 2017-02-09T18:35:52.990

5

when i==11 this loop is wrong:

for (j=i+1;x[j]!='#';j++){

since it makes j start at 12. You are already outside the string, but advance forward in the woods since your loop stops on #: undefined behaviour.

It stops if you encounter a # char while going through unallocated memory (can happen or not) or when you encounter a page boundary beyond what the system detects an illegal read and stops with SEGV.

edited Feb 09 '17 at 18:35

answered Feb 09 '17 at 13:13

Jean-François Fabre

131,796
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i see i see ... ty – domdrag Feb 09 '17 at 13:15
1

There is no null terminating character as the initializer are individual characters. – Paul Ogilvie Feb 09 '17 at 13:20
@PaulOgilvie: of course! the woods are closer than I thought (woods or bush depending on the hemisphere). – Jean-François Fabre Feb 09 '17 at 13:23
Is this error a segmentation fault right ? – Suraj Jain Feb 09 '17 at 15:24
yes, you end up reading somewhere where it's not allowed. – Jean-François Fabre Feb 09 '17 at 15:25
I have edited question to reflect this and some other error can you approve it. – Suraj Jain Feb 09 '17 at 15:27
Sir , if you have say example int x[11]; and you then compare x[11] < a , where a is some number, will program still crash , you are not writing on that location. ?? – Suraj Jain Feb 09 '17 at 15:36
it could crash but would be very unlikely. You have to hit a MMU boundary. The problem here is that `#` is not found so your loop continues until it finds one, or ... a MMU boundary: that's where it crashes. – Jean-François Fabre Feb 09 '17 at 15:48
I learned that comparing involve subtraction , so wont it also cause segmentation fault. and what is mmu ? Plus i see you joined 8 months ago, very very impressed 34000 in 8 months. – Suraj Jain Feb 09 '17 at 16:31
MMU is the device that protects against illegal writes but also _reads_. That's the problem here. Yeah 34000 is a lot of work :) – Jean-François Fabre Feb 09 '17 at 18:36

score 0 · Answer 2 · edited May 23 '17 at 12:24

What You Are Facing is Segmentation Fault.
You are accessing memory which you are not supposed to.

This Part Of Your Code:

for (i=0;i<12;i++){          //Till i<12, that means when i = 11 loop will run.
        if (x[i]=='#'){
            for (j=i+1;x[j]!='#';j++){  //j =  i+1, for i = 11, j = 12
                if (x[j]=='v'){         
                    v=v+1;
                }
                if (x[j]=='o')
                    o=o+1;
            }

            if (v>=o){
             for (j=i+1;x[j]!='#';j++){  //j =  i+1, for i = 11, j = 12
               if (x[j]=='o'){         //x[j] , for j = 12 , x[12]
                x[j]='.';  //x[12] = '.' You Are Accessing Memory You Are Not Supposed to.
                    }
                }
            }

You should not access array out of bound, it has many risks.
Please See This Question How dangerous is it to access an array out of bounds?, I highly recommend it.

Program Crashes - Segmentation Fault

2 Answers2