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Unsolved Segmentation Fault

This is a discussion on Unsolved Segmentation Fault within the C Programming forums, part of the General Programming Boards category; So, I'm getting a segmentation fault I can't solve. I have this code, with print statements to show where the ...

  1. #1
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    Unsolved Segmentation Fault

    So, I'm getting a segmentation fault I can't solve. I have this code, with print statements to show where the program quits.
    Code:
    i = 0;
    	while(i<2000){
    		printf("Host name %s : %i \n i is %i\n", hostCon[i].hostName, hostCon[i].freq, i);
    		fprintf(host_freq, "%i\n", hostCon[i].freq);
    		printf("hi1\n");
    		fprintf(host_name, "%s", hostCon[i].hostName);
    		printf("hi2\n");
    		i++;
    		}
    	printf("out of for loop");
    And the print looks like this:

    Host name *******.com
    : 1
    i is 1999
    hi1
    hi2
    Segmentation fault


    It seems odd to me, because after the "hi2" the while loop should be finished, thus the next print statement should I appear. I don't understand how this segmentation fault occurs when there is nothing in between the two print statements. Let me know if more code is necessary to debug this. Thanks in advance.

  2. #2
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    Seg fault is usually when you try to access memory that is out of bounds. Most commonly happens when you try to access an array past it's limit. My guess would be your while loop isn't terminating when you think it does. Hard to tell without more code.

  3. #3
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    What have you done to debug the code and narrow it down to the block that maybe causing the segfault.

  4. #4
    Algorithm Dissector iMalc's Avatar
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    There is nothing fatally wrong with that piece of code. Show more related code (definitions / allocations etc).
    stahta01 likes this.
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    Advice: Take only as directed - If symptoms persist, please see your debugger

    Linus Torvalds: "But it clearly is the only right way. The fact that everybody else does it some other way only means that they are wrong"

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    The print statements were added to see where the seg fault is. I'm having trouble with it, because the statements prints to show that i= 1999, which means it should quit the while loop after it is incremented and continue to the next print statement. I understand a seg fault has to do with accessing memory, but in this case there isn't any array or pointer that would cause a normal seg fault in between the "hi2" and "out of loop" print statements. Here is more code if it helps:

    Code:
    void sort(struct hostfreq hostCon[2001]){
    	FILE *host_freq;
        host_freq= fopen("hostfreq.txt", "a");
        FILE *host_name;
        host_name= fopen("hostname.txt", "a");
    	//printf("in sort\n");
    	struct hostfreq temp[1];
    	//printf("defined temp\n");
    	int i, j = 0;
    	for(i; i<=1999; i++){
    		j=i+1;
    		for(j; j<=1999;j++){
    				if(hostCon[j].freq>hostCon[i].freq){
    					temp[0] = hostCon[j];
    					hostCon[j] = hostCon[i];
    					hostCon[i]=temp[0];
    				}
    		}
    	}
    	i = 0;
    	while(i<2000){
    		printf("Host name %s : %i \n i is %i\n", hostCon[i].hostName, hostCon[i].freq, i);
    		fprintf(host_freq, "%i\n", hostCon[i].freq);
    		printf("hi1\n");
    		fprintf(host_name, "%s", hostCon[i].hostName);
    		printf("hi2\n");
    		i++;
    		}
    	printf("out of for loop");
    	fclose(host_freq);
    	fclose(host_name);
    	int total = sum(hostCon);
    	printf("sort done");
    	CDF(hostCon, total);
    }

  6. #6
    spurious conceit MK27's Avatar
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    Using printf to debug is prone to the following error: stdout is buffered, meaning you could have this:

    Code:
    printf("Everything okay");
    [...]
    [some line causing segfault]
    But when you run it, the message is still in the stdout buffer at the crash, making you believe that the printf line has not been executed yet when actually it has. So either:

    Code:
    printf("Everything okay"); 
    fflush(stdout);
    // or
    fprintf(stderr, "Everything okay");
    Since stderr is not buffered.

    By far the easiest way to locate a segfault is in a real debugger, which evidently you do not use, but from the look of your code, you are at a point where you do need to learn to ASAP. Let us know what OS you are using and someone can get you started in the right direction.

    Otherwise it's hard to say. Eg, you use fopen() without checking for errors, and using a bad handle will cause a segfault. Etc.
    C programming resources:
    GNU C Function and Macro Index -- glibc reference manual
    The C Book -- nice online learner guide
    Current ISO draft standard
    CCAN -- new CPAN like open source library repository
    3 (different) GNU debugger tutorials: #1 -- #2 -- #3
    cpwiki -- our wiki on sourceforge

  7. #7
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    Variable i is being used without being initialized.
    Code:
    for(i; i<=1999; i++){
    It's probably some out-of-bounds garbage.
    stahta01 likes this.

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    Also there's no check to see if the files actually opened or not...

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    MK27, thanks for the advice with using printf.
    My trouble is the past 3 years I've been programming in Java and for my current research they asked me to change to C, because it needs to be lower level coding. I'm used to using IDEs, but haven't made the move to one for C yet. Segmentation faults seem to be my nemesis since I started 3 weeks ago. I'm running Ubuntu and coding in Gedit. Is there a debugging plugin of some sort? I'd prefer to stay with a very simple program still, last time I coded in C or C++ was 3 years ago and it was on Microsoft Visual studio, which was the biggest headache ever. I used eclipse for java, but it still wasn't my favorite. If someone can still spot the seg fault above, it would be much appreciated though.

    nonoob, I believe i is initialized the line above "int i, j = 0"

  10. #10
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    No it isn't. Just j is initialized by that line. You need to do int i = 0, j = 0

  11. #11
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    Quote Originally Posted by uconnhuskies View Post
    nonoob, I believe i is initialized the line above "int i, j = 0"
    Actually with that code, j is initialized to 0 but i is not initialized at all and might contain any random value.

  12. #12
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    My fault, I changed it to initialize i also. Thanks.

    To check if a file is open do I just check that the stream is not equal to NULL?

  13. #13
    ATH0 quzah's Avatar
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    Quote Originally Posted by uconnhuskies View Post
    To check if a file is open do I just check that the stream is not equal to NULL?
    Yep.
    Code:
    fp = fopen( ... )
    if( fp == NULL )
        ...boo...
    else
        ...yay...

    Quzah.
    Hope is the first step on the road to disappointment.

  14. #14
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    Check the man pages:
    Quote Originally Posted by man fopen
    RETURN VALUE
    Upon successful completion fopen(), fdopen() and freopen() return a FILE pointer. Otherwise, NULL is returned and errno is set to indicate the error.

  15. #15
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    Quote Originally Posted by uconnhuskies View Post
    My fault, I changed it to initialize i also. Thanks.

    To check if a file is open do I just check that the stream is not equal to NULL?
    Yep... like this...

    Code:
    FILE *host_freq;
    host_freq= fopen("hostfreq.txt", "a");
    
    if (! host_freq)
     { printf("Host Frequency file not opened!")
        exit(1); }
    Obviously you don't want to continue processing if the files are not correctly opened.
    TheBigH likes this.

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