char *someFun1()
{
char temp[ ] = “string";
return temp;
}
char *someFun2()
{
char temp[ ] = {‘s’, ‘t’,’r’,’i’,’n’,’g’};
return temp;
}
int main()
{
puts(someFun1());
puts(someFun2());
}
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Answer Posted / rahulkulkarni
Variables are in function calls, hence on heap space. Variable address is being returned by function and when function returns the allocated space is freed.
Now accessing returned address will result in : segmentation fault.
As address no longer allocated to function, results in invalid address accessed by process.
As variable address does not exist ( hence variable address also , as heap is freed).
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