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c is a pointer to a read only location. It is in .text segment and not .bss since "abc" is a literal. Attempt to write there should crash the program.

For heaven's sake STOP! using TurboC.

Or the problem is its a pointer constant in line 1(char *c="abc" )which you try to move in line 2( *c='d'), not allowed in C

the program will crash cause the value of *c cannot be changed.

I think because you did not allocate memory to the pointer.

you cant change the value, it will throw an exception.

you can't change the value.

String literals cannot be changed in C. The behavior in this case is undefined. Your char* c is pointing to a string literal. It is a really good idea to define char* c as const char* c to catch this kind of error during compile time. Or initialize an array, rather than a pointer with the literal if you'll change it later, as in char c[10] = "abc".

c is a pointer to a read only location. It is in .text segment and not .bss since "abc" is a literal. Attempt to write there should crash the program.

It will lead to the crashing of the program ..

String literals are treated as constant strings, so, in summary you can't change them. If you run strings <binary> you'll see abc as such stored in the BSS section of the binary. So, any attempt to modify a constant memory location would result in an undefined behaviour or crash.

My TurboC compiler gives 'dbc' as output.

My TurboC compiler gives 'dbc' as output.

For heaven's sake STOP! using TurboC. The program should crash if compiled with a decent compiler.

what i think... U cant assign the val to pointer... as 'd' gives the ascii for d and assigning it,dosn't justify

your answer is correct..

Segmentation fault
*c = 'd';

segmentation fault