what is the difference between these initializations? Char a[]=”string”;

what is the difference between these initializations?
Char a[]=”string”;
Char *p=”literal”;
Does *p++ increment p, or what it points to?

Question Posted / a. sujatha

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what is the difference between these initializations? Char a[]=”string”; Char *p=”literal�..

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

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what is the difference between these initializations? Char a[]=”string”; Char *p=”literal�..

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.

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what is the difference between these initializations? Char a[]=”string”; Char *p=”literal�..

Answer / vignesh1988i

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

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what is the difference between these initializations? Char a[]=”string”; Char *p=”literal�..

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.

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