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

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Question

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

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

Answer
# 1

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 ?

0 Yes

1 No

Deepa

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

Answer
# 2

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 ?

1 Yes

1 No

Deepa

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

Answer
# 3

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

1 No

Anu

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

Answer
# 4

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 ?

0 Yes

0 No

Vijay

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

Answer
# 5

it prints BYE on screen!!!

Is This Answer Correct ?

2 Yes

0 No

Prasad

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

Answer
# 6

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 ?

0 Yes

2 No

Shruti

Question	Asked @	Answers
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