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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?
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Answer Posted / 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 |
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