Discussion :: Arrays
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Let x be an array. Which of the following operations are illegal?
I. ++x
II. x+1
III. x++
IV. x*2
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Answer : Option D
Explanation :
int x[10]; * x will store the base address of array. *
Statement I, III and IV is invalid.
Statement I and III : ++x and x++ are throwing en error while compile (lvalue required as increment operand )
Since, x is storing in the address of the array which is static value which cannot be change by the operand.
Statement IV : x*2 is also throw an error while compile (invalid operands to binary * (have 'int *' and 'int') )
Statement II : x+1 is throw a warning: assignment makes integer from pointer without a cast [enabled by default]
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