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> (i - (i -1))));
}
}
">main(){ signed int bit=512, i=5; for(;i;i--) { printf("%dn", bit = (bit >> (i - (i -1)))); }}
A) 128, 64, 32, 16, 8B) 256, 128, 64, 32, 16C) 512, 256, 128, 64, 32D) 64, 32, 16, 8, 4
sittoop
- Feb 26th, 2009
Right shift a number by 1 is equivalent to division of that number by 2^1.
Right shift a number by 2 is equivalent to division of that number by 2^2.Applying that, the answer is 256, 128, 64, 32, 16(left shift is opposite ~ multiplication)satyam_un
- Mar 13th, 2006
Ans is Bbecause the will b terminated after i=0;bit>>(i-(i-1)))means first it assignsi=5.the first value for bit is512>>(5-(5-1))=512>>1=512/2/1=256
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printf("%d, %d ", j++ , ++i);
}
">main(){ int i =10, j = 20; clrscr(); printf("%d, %d, ", j-- , --i); printf("%d, %d ", j++ , ++i);}
A) 20, 9, 19, 10B) 20, 9, 20, 10C) 20, 10, 20, 10D) 19, 9, 20, 10
abhimanipal
- Jan 28th, 2010
The O/P in this question is not compiler dependant. The language specification of C clearly states that the arguments to the function are supposed to be passed from right to left. So irrespective of the compiler there should be only 1 answer20, 9, 19, 10
sittoop
- Feb 26th, 2009
This is supposed to be same in both C and C++. According to ISO ANSII standard the answer to this is "undefined".
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}
">main(){ printf("%d, %d", sizeof('c'), sizeof(100));}
A) 4, 100B) 2, 100C) 2, 2D) 4, 4
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How will you print % character?
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p[0] = 'H';
printf("%s", p);
}
">main(){ char *p = "hello world"; p[0] = 'H'; printf("%s", p);}
A) Compile errorB) “Hello world”C) Runtime error.D) “hello world”
