...definition of "Segmentation Fault" - Where is...

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Answer by An
Submitted on 2/13/2005
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//n is a pointer to an array of nodes elements //heapArr is a pointer in the heap class to the array where //the heap nodes are stored bool heap::buildHeap(HeapNode *n, const int nodes) { //delete previous heap if one exists if(curSize != 0) { delete [] heapArr; heapArr = NULL; curSize = 0; maxSize = 0; } heapArr = new HeapNode[nodes]; //segmentation Fault if (heapArr != NULL) { maxSize = nodes; for (int a = 1; a <= maxSize; a++) { cout << "begin copy" <<endl; heapArr[a] = n[a]; curSize = a; percDown(a); } return true; } else maxSize = 0; return false; }

Answer by vicky
Submitted on 6/20/2005
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A segmentation fault occurs when your program tries to access memory locations that haven't been allocated for the program's use. Here are some common errors that will cause this problem: scanf("%d", number); In this case, number is integer. scanf() expects you to pass it the address of the variable you want to read an integer into. But, the writer has fogotten to use the `&' before number to give scanf the address of the variable. If the value of number happened to be 3, scanf() would try to access memory location 3, which is not accessible by normal users. The correct way to access the address of number would be to place a `&' (ampersand) before number: scanf("%d", &number); Another common segmentation fault occurs when you try to access an array index which is out of range. Let's say you set up an array of integers: int integers[80]; If, in your program, you try to use an index (the number within the brackets) over 79, you will ``step out of your memory bounds'', which causes a segmentation fault. To correct this, rethink your array bounds or the code that is using the array.

Answer by swarup
Submitted on 7/21/2005
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A segmentation fault occurs when your program tries to access memory locations that haven't been allocated for the program's use. Here are some common errors that will cause this problem: scanf("%d", number); In this case, number is integer. scanf() expects you to pass it the address of the variable you want to read an integer into. But, the writer has fogotten to use the `&' before number to give scanf the address of the variable. If the value of number happened to be 3, scanf() would try to access memory location 3, which is not accessible by normal users. The correct way to access the address of number would be to place a `&' (ampersand) before number: scanf("%d", &number); Another common segmentation fault occurs when you try to access an array index which is out of range. Let's say you set up an array of integers: int integers[80]; If, in your program, you try to use an index (the number within the brackets) over 79, you will ``step out of your memory bounds'', which causes a segmentation fault. To correct this, rethink your array bounds or the code that is using the array.

Answer by Valdo
Submitted on 11/15/2006
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A very simple c program: hd.c compiled with gcc -o hd hd.c run: ./hd <somefile> The idea is print the file contents as hexadecimal. but after print the first 16 bytes, a segmentatio fault occurs. I am use this code before in UNIX and DOS/Windows, and never got this error. What Am I doing wrong? TIA Valdo #include <stdio.h> #include <string.h> int main(int argc, char *argv) { FILE f; unsigned char fs[15]; int i=0; long offset=0; f = fopen(argv[1], "r"); fseek(f, 0, SEEK_SET); while (!feof(f)) { offset =+ fread(fs, 1, 16, f); printf("%010d ", (offset-16)); for (i=0; i < 16; i++) { printf("%02X", fs[i]); printf((i==7)?("-"):(" ")); } printf("\t"); for (i=0; i < 16; i++) { printf("%c", fs[i]); } printf("\r\n"); } return 0; }

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Question by SLayne
Submitted on 7/10/2003
Related FAQ: comp.unix.aix Frequently Asked Questions (Part 5 of 5)
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What is the definition of "Segmentation Fault" - Where is this defined?

Answer by David
Submitted on 7/16/2003
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Segmentation Fault occurred when running c program that is reading or writing non-exist segment(physical memory space). Example, declare an array of int x[5] and you try putting or reading 6th item(s) that the cpu system to crash.

Answer by Dragon Dave
Submitted on 8/25/2003
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It can be a sporadic occurance; for example, in the program below, if you type 'Dragon' it works quite happily. It's overwriting some part of the memory with something it shouldn't be. But it doesn't segfault until you get to something like 'Dragon Dave is overwriting your memory!'. This is a buffer overrun, and that's bad, and a very, very large hole in your programs security. #include "stdio.h" int main() { /* allocate 5 bytes for the string / char name[5]; / enter a string - possibly much longer / gets(name); / display the string */ printf("Hello, %s.",name); }

Answer by kon
Submitted on 10/22/2003
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segmentation faults may also occur in case of hardware errors, f.e. hard-disk or Memory failure. to find out, try running memtest86 and/or any hard-disk diagnostic tools from the manufacturer

Answer by prashant
Submitted on 11/4/2003
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segmentation fault is a type of error which occurs when u try to access a non existant physical memory address. Sometimes during execution of a C program u freed memory and then within the scope of the same program try again to use the same memory, then also the seg. fault occurs, this is due to the fact that untill the completion of the program the memory utilised is not returned to the operating system.

Answer by Liang Yu
Submitted on 12/11/2003
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This is often caused by improper usage of pointers in the source code, dereferencing a null pointer, or (in C) inadvertently using a non-pointer variable as a pointer.

Answer by Ryan
Submitted on 1/18/2004
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Segmentation fault is a bad error.

Answer by ashwin
Submitted on 2/14/2004
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if u allot some memory for array in the beginning of the programme and if u try to store the more numbers in the array and if it requires more memory that that has been allocated then it will result in segmentation fault

Answer by kilgroja
Submitted on 3/26/2004
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Ok. I understand. But...what about this: void memPtr = malloc(25); //causes coredump But, if I change the statement to: void memPtr = malloc(24); //ok Anything > 24 in this particular program causes a coredump. My debugger (ladebug) tells me that I got a signal segmentation fault. The system I'm running on is a Compaq Tru64 system with oodles of memory. I can port it to another Tru64 system and the same thing happens. Can anyone give me a clue what might be happening?

Answer by logicTRANCE
Submitted on 4/17/2004
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did you check on other machines? other than Compaq Tru64.... if it runs properly on other machines, report compaq call centre or equivalent...it must be a bug.... also, if you are using C++, try using void *memPtr = new(24); i've never tried it..but i think it should work... goodluck! Happy Programming!

Answer by Nothix
Submitted on 4/26/2004
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Thanks for the pointer Tip!

Answer by bt
Submitted on 5/4/2004
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get as point please

Answer by anand
Submitted on 6/19/2004
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This is often caused by improper usage of pointers in the source code, dereferencing a null pointer, or (in C) inadvertently using a non-pointer variable as a pointer.

Answer by dsff
Submitted on 10/29/2004
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address

Answer by harik
Submitted on 8/18/2005
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Pointer answer is a good one

Answer by rcw
Submitted on 12/26/2005
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Why the following will cause segmentation fault ? void calculatedouble() { double* d1; double* d2; double d; d1 = 2.3; d2 = 3.7; //segmentation fault here d = (d1) (*d2); cout<< d <<endl; }

Answer by RenjithLal
Submitted on 9/18/2005
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A segmentation fault occurs when your program tries to access memory locations that haven't been allocated for the program's use.

Answer by Nilesh bhosle
Submitted on 1/2/2006
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segmentation fault occurs due to 2/3 reasons.it may occur due to long array size,due to use of pointers in u r code.

Answer by siva reddy
Submitted on 1/23/2006
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if u allot some memory for array in the beginning of the programmer and if u try to store the more numbers in the array and if it requires more memory that that has been allocated then it will result in segmentation fault

Answer by nerdy_nerd
Submitted on 2/1/2006
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seg fault can occur when something goes wrong with the program that is using something which is not exist or faulty I would say that these kinds of error should be avoid at the begning of the program,otherwise it will cost a lot in the future. for example if you are writing a program for a big companies, then its too late to chase those seg fault errors. When writing the program, bear in mind that when you start always be consistent and focus on what you do. thats always a good practice. Hope this would give you some kind of thought in writing a well matured programs. If you have further doubts let me know. NerDy_NeRd

Answer by Prem
Submitted on 2/6/2006
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To solve this problem use gdb run ur code as gcc -ggdb -o obj file.c gdb r some_argu line no or use gdb bt go through that part of code and come out of this deadly problem

Answer by bulla
Submitted on 2/7/2006
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This is often caused by improper usage of pointers in the source code, dereferencing a null pointer, or (in C) inadvertently using a non-pointer variable as a pointer.

Answer by dude
Submitted on 3/12/2006
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sssuck on ma balls all of u

Answer by shrikant
Submitted on 5/29/2006
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segmentation fault is one tye of error occuerd during running the program its nothing but referencing illegal or invalid memory area.

Answer by Nitin
Submitted on 6/20/2006
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I am also getting same problem during malloc,what could be the solution for that?

Answer by bipin
Submitted on 8/2/2006
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well thhnx a lot 4 te ans on pointers

Answer by kumar
Submitted on 11/25/2006
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when we call a main function with in main, and in main we initialised two variables. this makes infinite loop.while running these variables get stored in different locations on memory. when there is no memory avialable to store these variables after some running time this gives segmentation fault. try one example.

Answer by santosh
Submitted on 12/27/2006
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#include<stdio.h> int main() { char *p = "srikanth"; p[3]='u'; printf("%s",p); } // here also u get seg fault // can anyone give me the reason

Answer by pintu
Submitted on 2/22/2007
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error is due to the address

Answer by Devagi_n
Submitted on 3/8/2007
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Segmentation fault occurs when a program attempts to access the memory location that is not allowed to access.

Answer by yuggi
Submitted on 3/11/2007
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an example for seg_fault ...... ...... scanf("%d", x); --------------- the routine scanf requires &x or for x to be a pointer you mentioned earlier .if this fails x does point to some a memory and if it is invalid you the the seg_fault. ~All pain yes gain ! ~

Answer by ituns
Submitted on 4/15/2007
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there are many soloution. If u care,pm to me. tuongtuong@mac.com

Answer by Apprentice
Submitted on 5/9/2007
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Hello! I have a "Segmentation fault" when i run xchat. Anybody can help me? Appreciate

Answer by harha b
Submitted on 5/14/2007
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any variable which is trying to acess memroy has which has been not allocated or trying to acess the freed memory will result in segmentation fault.

Answer by jhetfield18
Submitted on 5/14/2007
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Let's say I have an Array[6][6] and its integer.If we try to overwrite some of its content with a character '' and then try to print the array on the screen but with the '' wherever we've put it.Is it possible to get a segfault?

Answer by Panshul
Submitted on 6/5/2007
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segmentation fault occurs when a program attempts to access a memory location that it is not allowed to access, or attempts to access a memory location in a way it is not allowed to (eg, attempts to write a read-only location).