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Reason for Segmentation fault

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Top Forums Programming Reason for Segmentation fault
# 8  
Old 11-01-2007
Quote:
Originally Posted by royalibrahim
The following program fails with "Segmentation fault" error message, while I try to run in Ubuntu (Debian) Linux m/c. It is not creating any core file, so I could not cross examine it with the debugger. See the comments for much better understanding. Could any one tell me the exact reason why the program is failing?

Code:
int main( ) {
    char *ch; (or) ch = 'A';   // but if it assigned to any string then no segmentation fault
    int *p = (int*) &ch[0];   // or &ch[1], &ch[2], …. ch;      but &ch runs fine
    printf("%c", *p);           // Segmentation Fault: only if you use this print statement
}

You didn't allocate memory to ch.
Try:
Code:
int main( ) {
    char *ch = malloc(1); 
    *ch = 'A'; 
    int *p = (int*) &ch[0];   // or &ch[1], &ch[2], …. ch;      but &ch runs fine
    printf("%c", *p);           // Segmentation Fault: only if you use this print statement
}

# 9  
Old 11-21-2007
Just to sidetrack a bit, im aware that

Code:
char *ch;
*ch = 'A';

should be the correct way to do the ptr value assignment, however I did a bit of investigating on my own and gcc warned me of a type conversion error (something about from an int ptr to a char)

However, if I do

Code:
char *ch;
*ch = "A";

gcc is happy

Is it just me, or you cannot assign char ptrs to a single quote character ?
# 10  
Old 11-21-2007
Quote:
Originally Posted by JamesGoh
Is it just me, or you cannot assign char ptrs to a single quote character ?
You need memory allocated to ch, any of the following are valid..


Code:
char *ch=malloc(1);
*ch='A';

Code:
char *ch=malloc(1);
ch[0]='A';

Code:
const char *ch="A";

Code:
char ach[1];
char *ch=ach;
ch[0]='A';

you need to understand (a) memory (b) pointers

otherwise move to directly to Java.
# 11  
Old 11-21-2007
Found out another way to do it which sticks strictly to the theory behind pointers

Code:
char *ptr;
char c='a';
ptr = &c;

Please feel free correct me if I am wrong, as I want to show that I properly understand pointer theory
# 12  
Old 11-21-2007
Quote:
Originally Posted by porter
You need memory allocated to ch, any of the following are valid..


Code:
char *ch=malloc(1);
*ch='A';

Code:
char *ch=malloc(1);
ch[0]='A';

Code:
const char *ch="A";

Code:
char ach[1];
char *ch=ach;
ch[0]='A';

you need to understand (a) memory (b) pointers

otherwise move to directly to Java.
Hi,

Agree with what Porter state, it is a good practise to allocate a memory block by using malloc rather than directly referencing the pointer to a hardcoded value, this will created problem during run time althought the code can be compiled.

Cheers
# 13  
Old 11-22-2007
By the way, many Linux distributions disable core file creations by default. Stupid for developers, but safer for users (I guess.)

Try:
Code:
ulimit -c unlimited

Before running your program, of course Smilie You should then get a core on a SIGSEGV.
# 14  
Old 12-06-2007
The method I suggested for assigning characters to pointers

Code:
char *ptr;
char c='a';
ptr = &c;

compiled and ran fine . So if its working properly, why should I still use the malloc() and free() method to assign and deallocate pointer memory ? Is this due to the greater control one is given in manipulating the memory from the heap ?
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