pop and peek doing same in that case?

Question

From all the sources I've read, they say - the difference between peek and pop is that peek doesn't remove the top value. In the provided example from my lecture notes, apparently they do the same using a different method of subtraction. After both operations top has 1 subtracted.

Am I right? Probably not, can somebody explain how do those differ?

int pop(void)
{
    assert(top>0);
    return data[--top];
}
int peek(void)
{
    assert(top>0);
    return data[top-1];
}

pb2q · Accepted Answer · 2021-07-05T18:27:53.130

top is a state variable for your stack, and in this case it is a stack which is stored in a regular array. The variable top references the top of the stack by storing an array index.

The first operation, pop, uses the decrement operator to change the variable top. --top is equivalent to top = top - 1. And hence after a call to pop the state of the stack is also changed. The old value will remain in the array, but since the variable top now references a different index, this value is effectively removed: the top of the stack is now a different element. Now if you call a push, this popped value will be overwritten.

The second operation doesn't change the value of the variable top, it only uses it to return the value at the top of the stack. The variable top still references the same value as the stack's top, and so the stack is unchanged.

score 5 · Answer 2 · answered Sep 14 '12 at 19:57

5

They will return the same value but only pop changes top:

--top

this is equivalent to

top = top -1

where as top - 1 does not change the value of top.

answered Sep 14 '12 at 19:57

Borgleader

15,826
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score 5 · Answer 3 · answered Sep 14 '12 at 20:04

In general programming terms, "pop" means the method of returning an object from a stack, while at the same time removing it from the stack. The term "peek" is more generic and can be used on other data containers/ADTs than stacks. "Peek" always means "give me the next item but do not remove it from the container".

Most often "peek" is used together with queue-like data containers, for example the Windows API function for checking the next message in the windows message queue is named PeekMessage().

score 2 · Answer 4 · answered Sep 14 '12 at 19:58

This

return data[--top];

changes the value of top while this

return data[top-1];

does not change it. Thus, when you call pop the pointer to the top of the stack is modified to point to a new item while peek leaves the pointer unchanged so the functions behave in the correct sense you describe

score 2 · Answer 5 · answered Sep 14 '12 at 19:58

2

In pop its doing --top which is top=top-1 so it changes the value of top.

While in peek, its doing top-1 i.e. just decrements top by 1 and use the value, top is unchanged.

answered Sep 14 '12 at 19:58

Rohan

52,392
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score 0 · Answer 6 · answered Jan 23 '19 at 05:32

Function of pop - It extracts the top element and move to the next before it.so the pointer now moved to the next element.top element position is now decrease by 1.

Function of peek - It only return the top element but the pointer is still there. so the top element position remain unchanged.

score -2 · Answer 7 · answered Apr 09 '16 at 16:34

-2

actually pop and peak are same but the thing is the former saw the data and closed the box but the later first took the data saw it and closed the box

answered Apr 09 '16 at 16:34

faizy

1

pop and peek doing same in that case?

7 Answers7