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what is the difference between these initializations?
Char a[]=”string”;
Char *p=”literal”;
Does *p++ increment p, or what it points to?
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Answer / bee
logically, both are treated as array of characters(i.e.
string) but....
1) a is an array of characters(a string)
2) p is a pointer to an array of characters
the statement char *p = "literal" is equivalent to
char j[] = "literal"
char *p = j;
3) *p++ can be seen as *(p++)....
this is so because '++' has higher recedence over '*'
operator. so, it increments address by 1 unit and prints
the corresponding value value
| Is This Answer Correct ? | 5 Yes | 0 No |
Answer / gaurav
I am totally satisfied with your above explanation except
last one.
i.e. Char *p="literal";
So, i want to mention yes this will work.
Explanation: *p++.
Here we have post increment.
Postfix increment/decrement have high precedence, but the
actual increment or decrement of the operand is delayed (to
be accomplished sometime before the statement completes
execution).
value of printf("\nstr=%c\n",*p++) will be 'l', but before
complete execution of this statement p will point to string
"iteral" as p got incremented.
| Is This Answer Correct ? | 3 Yes | 1 No |
surely there is some difference.....
here 'a' is represented as array in which string gets stored
in consecutive locations......
p is a pointer variable where string is initilized... so in
p the base address of "literal " will get stored......
*p++ increments 'p' , but pertaining to some conditions.....
++ has more precedence than * , so first it will increment
the address and correspondingly it will show the value as *
precedes..... so after the increment the p points to 'i'...
thank u
| Is This Answer Correct ? | 5 Yes | 5 No |
Answer / koti
Actually char a[ ]="string" in this scenario string constant is stored in read-only memory section and also stack section.in this case you can modify the string constant.that modifications are happened in stack section .so here
*a ++ men's
1 ) a is pointing to base address of string constant .
2 ) *a men's inside content that is 's'.
3 ) *a ++ men's incrementing the asci value of 's'. After that you can print this array like
Printf("%s",a);
O/P : ttring.
Coming to the *p ="literal" this scenario
1 ) *p is stored in stack section why because it is auto variable.
2 ) "literal" this string constant is stored in read-only memory section.
3 ) P is pointing to string constant Base addres
Here *p++ men's you are training to change read only memory section contact. so it is an error why because
You can't modified the read-only memory content.
Main difference is using arrays string constant is stored in both stack and read-only memory section.
Using pointers string constant is stored in read-only memory section only .
Thank you.
| Is This Answer Correct ? | 0 Yes | 0 No |
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