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f(char *p)
{
p=(char *)malloc(sizeof(6));
strcpy(p,"HELLO");
}
main()
{
char *p="BYE";
f(p)
printf("%s",p);
}
what is the output?

Question Posted / guest

Answers were Sorted based on User's Feedback



f(char *p) { p=(char *)malloc(sizeof(6)); strcpy(p,"HELLO"); } main() { char *p=..

Answer / prasad

it prints BYE on screen!!!

Is This Answer Correct ?    8 Yes 0 No

f(char *p) { p=(char *)malloc(sizeof(6)); strcpy(p,"HELLO"); } main() { char *p=..

Answer / hussain reddy

BYE

Is This Answer Correct ?    4 Yes 0 No

f(char *p) { p=(char *)malloc(sizeof(6)); strcpy(p,"HELLO"); } main() { char *p=..

Answer / deepa

SOORY FOR POSTIN THE WRONG ANSWER THE ANSER WUD BE BYE COZ
THE *P DIES IN THE FUNCTION ITSELF AS WE ARE NOT RETURNING
THE STRING BACK IN THE MAIN PROGRAM

Is This Answer Correct ?    4 Yes 1 No

f(char *p) { p=(char *)malloc(sizeof(6)); strcpy(p,"HELLO"); } main() { char *p=..

Answer / vijay

The out put is "BYE".
Because the pointer p dies when function exit with out
return,in main pointer p points to only "BYE",so prinf
prints which p points in main.

Is This Answer Correct ?    2 Yes 1 No

f(char *p) { p=(char *)malloc(sizeof(6)); strcpy(p,"HELLO"); } main() { char *p=..

Answer / smirnov_amm

It will show "HELL"
the f function reserve only 4 bytes with the sizeof(6). it will return the sizeof (int) =4 bytes.
So you only allocate 4 bytes.
When performing the strcpy, you will corupt memory because you'll copy more than allocate.

Cheers

Is This Answer Correct ?    0 Yes 0 No

f(char *p) { p=(char *)malloc(sizeof(6)); strcpy(p,"HELLO"); } main() { char *p=..

Answer / yathish m yadav

the output is "hello".
here we are overwriting pointer *p thrice.
that is in the function we get an piece of memory from
malloc and assigned to p,
in the statement strcpy(p,"hello");
the malloc memory is lost and the compiler creates an char
array and copies the string "hello" and it makes the
character array as constant.

Is This Answer Correct ?    0 Yes 2 No

f(char *p) { p=(char *)malloc(sizeof(6)); strcpy(p,"HELLO"); } main() { char *p=..

Answer / deepa

the outpu is hello coz the pointer is pointing to a
location where the string "bye" is written in the next
program yu are using the same pointer to point to some
other string so it gets overwritten

Is This Answer Correct ?    2 Yes 5 No

f(char *p) { p=(char *)malloc(sizeof(6)); strcpy(p,"HELLO"); } main() { char *p=..

Answer / anu

the output wll be hello , since the argument is overwritten
with new memory in the function.

so in main, when it comes to printf, p points to the
allocatd memory, which contains hello

Is This Answer Correct ?    0 Yes 3 No

f(char *p) { p=(char *)malloc(sizeof(6)); strcpy(p,"HELLO"); } main() { char *p=..

Answer / shruti

The output would be "HELLO"..

though we are not returning the string, we are making
direct changes at the memory location..

so "bye" will be overwritten with "HELLO"


because we are using pointers, the dying pointer scenario
is not applicabe here..

Its a pointer, not a variable..


This function will work similar to -> swapping two numbers
using pointers..
juss check that prog if you fnd somewhere.. :-)
you will get the logic... :-)


Cheers...


--By the way a gud ques.. :-)

Is This Answer Correct ?    1 Yes 4 No

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