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What will be the output of the program? #include<stdio.h> #include<string.h> int main() { char s; char fun(); s = fun(); printf("%s\n", s); return 0; } char fun() { char buffer[30]; strcpy(buffer, "RAM"); return (buffer); } ?->(Show Answer!)
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1. What will be the output of the program? #include<stdio.h> #include<string.h> int main() { char s; char fun(); s = fun(); printf("%s\n", s); return 0; } char fun() { char buffer[30]; strcpy(buffer, "RAM"); return (buffer); }





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  • By: guest on 01 Jun 2017 06.02 pm
    The output is unpredictable since buffer is an auto array and will die when the control go back to main. Thus s will be pointing to an array , which not exists.
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