C formatted Output using printf( Confused)

[rufi]

hi group,

I am new to C and using gcc compiler in Linux.

I am having the following problems when using the printf() funtion to have the formatted output.

Here r some things which i m trying :

The range of short int(which is signed by default) is -32768 to + 32767.

short i = 32767;
printf(" %d\n" , i);
It prints 32767. Good. At this point i was assuming that if I give variable i, the value greater than its range, it won't accept and it will give me an overflow message.

But when i assigned i(type short) the value 32768 or +32768 which is 1 more than its range, it didn't give me an error , rather it gave me the value with a -sign. -32768. And when i print it with %u conversion specifier (to omit the -ve sign)in printf , it gives me strange output which is 4294934528.

And when i assign the value 32769(2 more than the range of short) to i with the %d specifer, it goes in minus as well as the value changes to - 32767.

I really have no idea what is happening and why data types like short, int, accept more ranges than they are supposed to.

Please if anybody can clear these confusions or give me the web-site where i can read some clear explanations about this. I m referring K&R(second edition) and it does not give any clear explanation about this.

thanks.

[ravenspoint]

Do you not get a compiler warning?

MSVC gives a warning on:

    short  s;
    s = 32769;
    printf("%d\n",s);
 

[ilikepython]

hi group,

I am new to C and using gcc compiler in Linux.

I am having the following problems when using the printf() funtion to have the formatted output.

Here r some things which i m trying :

The range of short int(which is signed by default) is -32768 to + 32767.

short i = 32767;
printf(" %d\n" , i);
It prints 32767. Good. At this point i was assuming that if I give variable i, the value greater than its range, it won't accept and it will give me an overflow message.

But when i assigned i(type short) the value 32768 or +32768 which is 1 more than its range, it didn't give me an error , rather it gave me the value with a -sign. -32768. And when i print it with %u conversion specifier (to omit the -ve sign)in printf , it gives me strange output which is 4294934528.

And when i assign the value 32769(2 more than the range of short) to i with the %d specifer, it goes in minus as well as the value changes to - 32767.

I really have no idea what is happening and why data types like short, int, accept more ranges than they are supposed to.

Please if anybody can clear these confusions or give me the web-site where i can read some clear explanations about this. I m referring K&R(second edition) and it does not give any clear explanation about this.

thanks.

The variable overflows which means it goes back to its minimum value, which in your case is -32768:

short x = 32767;
x++;                             // x = 32768
x++;                             // x = -32768
x++;                             // x = -32767
x++;                             // x = -32766