String Handling

char *t1 = malloc(sizeof(char)*10);
char *t2 = malloc(sizeof(char)*10);
writing above statements is not ANSI standard and is wrong to get desired result because malloc returns void* (pointer to void mean it returns pointer that points to nothing)

you have to parse the pointer returned by malloc as given below

char *t1 =(char*) malloc(sizeof(char)*10);
char *t2 = (char*)malloc(sizeof(char)*10);


sencondly, t1 and t2 do not hold values, t1 and t2 hold memory address as allocated by malloc so the statement t1=t2 does not assigns the values pointed by t2 but it assigns the address of the location pointed by t2 and after that statement t1 and t2 will hold same address or they will point to the same memory location. if you want to copy values
*t1 = *t2; is the statement.

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More Useful Topics...

 

History Of C..

In the beginning was Charles Babbage and his Analytical Engine, a machine
he built in 1822 that could be programmed to carry out different computations.
Move forward more than 100 years, where the U.S. government in
1942 used concepts from Babbage’s engine to create the ENIAC, the first
modern computer.
Meanwhile, over at the AT&T Bell Labs, in 1972 Dennis Ritchie was working
with two languages: B (for Bell) and BCPL (Basic Combined Programming
Language). Inspired by Pascal, Mr. Ritchie developed the C programming
language.

My 1st Program...


#include
#include
void main ()
{
clrscr ();
printf ("\n\n\n\n");
printf ("\t\t\t*******Pankaj *******\n");
printf ("\t\t\t********************************\n");
printf ("\t\t\t\"Life is Good...\"\n");
printf ("\t\t\t********************************");
getch ();
}

Next Step...


#include
#include

void main ()
{
clrscr ();
printf ("\n\n\n\n\n\n\n\n");
printf ("\t\t\t --------------------------- \n\n");

printf ("\t\t\t | IGCT, Info Computers, INDIA | \n\n");
printf ("\t\t\t --------------------------- ");

getch ();

}

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