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#include<stdio.h>
main()
{
char *p1;
char *p2;
p1=(char *) malloc(25);
p2=(char *) malloc(25);
strcpy(p1,"Ramco");
strcpy(p2,"Systems");
strcat(p1,p2);
printf("%s",p1);
}

Tell me the output?

Question Posted / naveed

Answers were Sorted based on User's Feedback



#include<stdio.h> main() { char *p1; char *p2; p1=(char *) malloc(25); p2=(char *) m..

Answer / fazlur rahaman naik

The output will b : RamcoSystems

Is This Answer Correct ?    17 Yes 0 No

#include<stdio.h> main() { char *p1; char *p2; p1=(char *) malloc(25); p2=(char *) m..

Answer / sumant

the output will be RamcoSystems
but we need 2 more libraries
#include<string.h> and
#include<alloc.h>
to run this program. in else case it will not work.

Is This Answer Correct ?    6 Yes 0 No

#include<stdio.h> main() { char *p1; char *p2; p1=(char *) malloc(25); p2=(char *) m..

Answer / mage

The program is not correct. What is present in memory
beyond "Ramco" is not known and we are trying to
attach "Systems". May be we are overwriting something which
is unsafe.

To concatenate two strings declare the first as array.

example: char p1[50];
char *p2;
p2 = malloc(25);
strcpy(p1, "Ramco");
strcpy(p2,"Systems");
strcat(p1,p2);

Is This Answer Correct ?    7 Yes 1 No

#include<stdio.h> main() { char *p1; char *p2; p1=(char *) malloc(25); p2=(char *) m..

Answer / v.srinivasan

#include<stdio.h>

main()
{
char *p1,*p2;
p1 = (char *)malloc(25);
p2 = (char *)malloc(25);
strcpy(p1,"Ramco");
strcpy(p2,"Systems");
strcat(p1,p2);
printf("%s",p1);
}



the output will be RamcoSystems

we don't need the following libraries under Linux 2.6
#include<string.h> and
#include<alloc.h>
to run this program.

Is This Answer Correct ?    5 Yes 0 No

#include<stdio.h> main() { char *p1; char *p2; p1=(char *) malloc(25); p2=(char *) m..

Answer / khaja

ramco system

Is This Answer Correct ?    3 Yes 1 No

#include<stdio.h> main() { char *p1; char *p2; p1=(char *) malloc(25); p2=(char *) m..

Answer / mastan vali.shaik

25Ramco25Systems

Is This Answer Correct ?    2 Yes 13 No

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