problem during typecasting during function calling

Question

In the program i am converting the character datatype into integer but it is displaying the ASCII code of the character, why? what should i do to print 4 as output?

void fun(int x)
{
cout<<x;
}

int main()
{
char ch='4';
fun((int)ch);
return 0;
}

I have tried changing the parameter from 'int x' to 'char x' and then typecasting in cout as 'cout<<(int)x;'

D-RAJ · Accepted Answer · 2021-02-09T12:07:15.990

0

Casting a char to int will do what it should, cast the stored value into an integer. When you say char ch='4';, the variable holds the ASCII value of '4', it does not store the integer value 4. So casting it will not give you the integer value 4.

To get the integer value of ch (get 4 from '4'), we can subtract the integer value by '0' so that we get the actual integer value,

void fun(int x)
{
    cout << x - '0';
}

edited Feb 09 '21 at 12:07

answered Feb 09 '21 at 11:57

D-RAJ

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The `atoi` has undefined behaviour since `&ch` is not a null-terminated string. The correct method is `ch - '0'`. – molbdnilo Feb 09 '21 at 12:01
Don't use `48` - it's neither standard nor portable. Only `'0'` is guaranteed to work everywhere and forever. – molbdnilo Feb 09 '21 at 12:04
@molbdnilo Noted! – D-RAJ Feb 09 '21 at 12:05
The number printed will always be an integer, since the result of the subtraction is an `int`, not a `char`. – molbdnilo Feb 09 '21 at 12:06

problem during typecasting during function calling

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