Writing a ceiling function in C

Posted on: May 9, 2018, by :

Writing a ceiling function in C

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The modulus operator (%) essentially gives you the remainder. That is, a % b will give you the remainder of dividing a by b (so 7 % 2 = 1, 10 % 4 = 2, etc.)

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submitted4 years agobybaltimore94

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Im trying to write a function that (kind of) emulates the ceiling function in the math library, in a way that is more useful for my purposes. The code Ive written is not working as expected. It should take two integer arguments, divide one by the other, and round up to the nearest integer no matter what the result is.

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To get floating-point division, you need to cast at least one of the two variablesnumberandvaluetofloatbefore you do the division. For example,

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ceiling(10,3); returns 3, when it should return 4.

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Thanks very much for your response. Makes sense now. Im not 100% sure but I dont think for this assignment Im allowed to use libraries other than stdio.h and stdlib.h. I have the rest of my program working fine (I think), it was just this one part giving me trouble, causing the entire program not to work as it should.

The variablesnumberandvalueare bothints. Therefore the expressionnumber/valueis integer division: it is the division of twointvalues. This is integer division, and it returns an integer value, regardless of whether you subsequently decide to store that integer value in afloatvariable or anintvariable.

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Ex: 2/2=1 — answer is 1; 10/3=3.333… — round up to 4.

For what its worth, why arent you just doing this?

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Wow, I dont know why it never occurred to me that using modulus would work. I even thought about it and passed it off thinking there wasnt a way I could employ it usefully. I must not have been thinking clearly, I had just woken up. Oh well, my program is finished and working properly now.

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Ninja edit: Now I also know what casting is and how to do it, so this worked out for me anyway.

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If youre not familiar with theternary operator, the above code is equivalent to:

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Here,numberis cast to afloatbefore the division. Because one of the operands of the division is a floating-point value now, the division is floating-point division, not integer division.

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ceiling(2,2); would return 1, as expected.

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A couple of other things, if you didnt know them already.

For what its worth, you can also utilize integer division in a ceiling function.

int ceiling(int n, int v) if(!n%v) // If v evenly divides into n, n/v is the same as ceil(n, v), return n/v; // so just return n/v else // Else, we know that integer division floors, so simply return (n/v) + 1; // return the floor plus one.

Writing a ceiling function in C(self.learnprogramming)

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My thought process is that if a is a float, then 10/3 should be 3.33333…, and if b is an int, then 10/3 should be 3. If c is a float, a-b should be 0.33333…, then since c!=0, ceiling should increment b and return that value. Where am I going wrong?

I use the!n%v, which is basically a shorter way of writingn%v == 0. In C, anifstatement evaluates to true for any non-zero value, and false only for a zero value.

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