Local variables Vs Global variable

Question

int main()
{
    int a;
    printf("the value is %d", a+'a');
    return 0;
}

In the above code a is local variable, And local variable are initialize to garbage value if we don't explicitly give them value . So the output should be some garbage value . But why am I getting output as 97?

You're invoking undefined behaviour. There might very well be stored some random value in memory where you variable currently points to. — lord.garbage, Jun 30 '15 at 12:17
Who knows what information is stored at a's memory address? In this case, it's 97. Junk value. — Matthew Darnell, Jun 30 '15 at 12:17
97 is actually a quite suspicious value, as it's the ASCII value for `a`. It could be that the compiler in this case, in a debug build, put the variable's name in there on purpose. Standardswise the value is of course undefined. — Niels Keurentjes, Jun 30 '15 at 12:19
@NielsKeurentjes more likely `a` happened to be `0`, and doing `'a' + 0` ends up with `'a'` — M.M, Jun 30 '15 at 12:20
nice to see everyone is up to date with C11 :) In C99 this was unspecified behaviour, if we are known to be on a system without trap representations. — M.M, Jun 30 '15 at 12:22
@NielsKeurentjes yeah i am confused here , that garbage value exactly matches with ASCII value . — p0s0731, Jun 30 '15 at 12:24
@NielsKeurentjes The variable name should be stored for the debug build, and it's very likely the reason. — Iharob Al Asimi, Jun 30 '15 at 12:26
Oh crap I missed the `+'a'` part, so it's just initialized to 0, either by accident or debugger. Will still be undefined on release build depending on the OS (`0xbaadfood` anyone?). — Niels Keurentjes, Jun 30 '15 at 13:05

score 2 · Answer 1 · answered Jun 30 '15 at 12:16

2

In your code,

 printf("the value is %d", a+'a');

produces undefined behaviour. The output of UB, is, well, undefined.

You cannot rely upon (or justify) the outcome (if any) for a statement which invokes UB.

answered Jun 30 '15 at 12:16

Sourav Ghosh

133,132
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score 0 · Answer 2 · answered Jun 30 '15 at 12:17

0

Local variables are stack variables. They are not initialized (unlike static variables). So better initialize yourself.

answered Jun 30 '15 at 12:17

Paul Praet

1,367
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3

This should be a comment. – Iharob Al Asimi Jun 30 '15 at 12:18

Local variables Vs Global variable

2 Answers2