The expression n++ requires a single machine instruction such as INR to carry out the increment operation whereas, n+1 requires more instructions to carry out this operation.

The question is meaningless since the two aren't equivalent.

n++ modifies n.
n+1 does not.

If the purpose is to increment n by one and n is a register variable, the two are almost the same in processor resources, except that the result of n+1 requires the compiler to allocate a second register for the intermediate result.

Also, the first answer is incorrect. The 'INC EAX' instruction in an x86 CPU is no faster than an 'ADD EAX,1' instruction and the only benefit is to reduce code space.

Memory related operations takes more time in m+1 it goes to the memory twice ie to search and to store whereas m++ goes only once

The expression n++ requires a single machine instruction such as INR to carry out the increment operation whereas, n+1 requires more instructions to carry out this operation.



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The question is meaningless since the two aren't equivalent.

n++ modifies n.
n+1 does not.

Well, it does not!
While we might be tempted to say that 'increment' instruction executes faster than a 'sum' instruction, I have not come across any compiler which uses 'sum' instruction for 'k+1'.
Also, this issue is totally dependent upon the instruction set of the machine (and compiler) for which the code is compiled. So, there is no point in making it a general rule.

n++ is unary operation and n+1 is binary operation.unary operation takes less time than binary operation.that is why n++ is faster than n+1

n++ requires less memory as compared to n+1 as well as n++ requires just 1 reg bt n+1 requires one more register to store the answer

hi..
n++
and n+=1 both will give the same results but when you compile these instructions seperately, compliler will generate only INR for n++ and
for n += 1 will generate ADD instruction . So this INR instruction is run little smaller than ADD.

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