C programming

Question

int main() {
		int a, b;
    /* Some code which initializes "a" here ... */
    if (a < 0) {
        a = -a;
    }
    b = sqrt (a);
}

• Above code will always work for calculating the square root of any valid integer value on  a system . [T/F]

Comments

how will it give correct answer for any non perfect square?

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by default sqrt() is of type double.

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Neglect b truncation. Other than that

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b I'll be integer. That's obvious. You can choose appreciate type for b, and make sqrt result as precise as possible.

Other than this problem ?

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no. No problem for any valid integer on system.

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@Debashish_deka but  the validity of this statement will depend on compiler   right ? Will we able to get a general answer for this ? I really doubt this get executed correctly always. Moreover for large -ve numbers , will it work ?

Answer 1

    #include <stdio.h>
     
    int main() {
    		int a=-4, b;
        /* Some code which initializes "a" here ... */
        if (a < 0) {
            a = -a;
        }
        b = sqrt (a);
        printf("%d",b);
     
    	return 0;
    }

yes code will always work. But will not give correct output everytime. Like for - 4 it is giving square root as 2. But it is not a correct result

Comments

^What has 32768 to do with int?

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Sir, I mean int range in between -32768 to 32767

But but output here showing 32768.

that is why it fails in INT_MIN

rt?

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int range in C is never that - that was in stone age. C standard does not fix the size of int, but most compilers take it as 4 bytes. So, range is $-2^{31} - (2^{31} - 1)$ if system uses 2's complement representation (this is also not mandated by C) and since there is one extra number possible for negatives,  it fails for INT_MIN.