CODEWITHSUNDEEP
clinuxrandom
Thair Abdalla
Thair Abdalla

Reputation: 279

Generating 10 bit signed integer

this line printf ("Random[%d]= %u\n", i, rand ()); generates a random integer mostly 1 bit, but when doing printf ("Random[%d]= %u\n", i, rand ()%1000000); it generates 6 digits int,

when trying to generate a 10 digit int the error " warning: format ‘%u’ expects argument of type ‘unsigned int’, but argument 3 has type ‘long long int’ [-Wformat]" occurs.

so is there a way to generate a random 10 bit signed integer???

Upvotes: 0

Views: 2581

Answers (2)

Some programmer dude
Some programmer dude

Reputation: 409166

You would have passed sa "long long" or "unsigned long long" either explicitly or implicitly to the printf, which is why you get the warning -- but more on that later;

Generating a 10 digit random number, you will need to switch to 64 bit -- so long long;

unsigned long long randomvalue;
randomvalue = random();
randomvalue <<= 16; // just picked 16 at random
randomvalue ^= random();  // you could also use + but not "or";
randomvalue %= 10000000000ULL;

The warning you get is because you pass a value of type long long as the third argument to the printf function, but you still use "%u" as the format, which makes printf expect a value of type unsigned int.

To silence the warning, change the format to "%lld".

It won't help you create a ten-digit number higher than 2147483647 though, as rand returns an int which so far is at most 32 bits, even on 64-bit platforms. Where the value 2147483647 comes from? The highest positive value if a signed 32-bit integer.

Upvotes: 1

SJuan76
SJuan76

Reputation: 24875

It does not fit in an int32 (check your platform to see how int is defined)

In an unsigned int32, the máximum value is 2^32 = 4,294,967,296. So when you do rand % 10000000000, 10000000000 is a long so % is computed as a long operation (which gives a result of such a type)

Upvotes: 1

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