CODEWITHSUNDEEP
ccasting
Rabiraj
Rabiraj

Reputation: 11

what happens when we type cast from lower datatype to higher datatype

Will the accessibility of memory space get changed or just informing the compiler take the variable of mentioned type?

Example:

int main()
{
    char a;
    a = 123456789;
    printf("ans is %d\n",(int)a);
}

Output:

overflow in implicit constant conversion a= 123456789.
ans is 21.

Here I know why it's causing overflow. But I want to know how memory is accessed when an overflow occurs.

Upvotes: 0

Views: 739

Answers (1)

Some programmer dude
Some programmer dude

Reputation: 409206

This is kind of simple: Since char typically only holds one byte, only a single byte of 123456789 will be copied to a. Exactly how depends on if char is signed or unsigned (it's implementation-specific which one it is). For the exact details see e.g. this integer conversion reference.

What typically happens (I haven't seen any compiler do any different) is that the last byte of the value is copied, unmodified, into a.

For 123456789, if you view the hexadecimal representation of the value it will be 0x75bcd15. Here you can easily see that the last byte is 0x15 which is 21 in decimal.

What happens with the casting to int when you print the value is actually nothing that wouldn't happen anyway... When using variable-argument functions like printf values of a smaller type than int will be promoted to an int. Your printf call is exactly equal to

printf("ans is %d\n",a);

Upvotes: 1

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