What is the difference between pre-increment and post-increment in the cycle (for/while)?

Question

My interest is in the difference between for and while loops. I know that the post-increment value is used and then incremented and the operation returns a constant pre-increment.

while (true) {
    //...
    i++;
    int j = i;
}

Here, will j contain the old i or the post-incremented i at the end of the loop?

Answer 1

Since the statement i++ ends at the ; in your example, it makes no difference whether you use pre- or post-increment.

The difference arises when you utilize the result:

int j = i++; // i will contain i_old + 1, j will contain the i_old.

Vs:

int j = ++i; // i and j will both contain i_old + 1.

Answer 2

Depends on how you use them.

• i++ makes a copy, increases i, and returns the copy (old value).
• ++i increases i, and returns i.

In your example it is all about speed. ++i will be the faster than i++ since it doesn't make a copy.

However a compiler will probably optimize it away since you are not storing the returned value from the increment operator in your example, but this is only possible for fundamental types like a int .

Answer 3

Basic answer for understanding. The incrementation operator works like this:

// ++i
function pre_increment(i) {
    i += 1;
    return i;
}
// i++
function post_increment(i) {
    copy = i;
    i += 1;
    return copy;
}

A good compiler will automatically replace i++ with ++i when it detect that the returned value will not be used.

Answer 4

In pre-increment the initial value is first incremented and then used inside the expression.

a = ++i;

In this example suppose the value of variable i is 5. Then value of variable a will be 6 because the value of i gets modified before using it in a expression.

In post-increment value is first used in a expression and then incremented.

a = i++;

In this example suppose the value of variable i is 5. Then value of variable a will be 5 because value of i gets incremented only after assigning the value 5 to a .

Answer 5

#include <stdlib.h>
#include <stdio.h>

int main(int argc, char **argp)
{
    int x = 5;

    printf("x=%d\n", ++x);
    printf("x=%d\n", x++);
    printf("x=%d\n", x);

    return EXIT_SUCCESS;
}

Program Output:

x=6
x=6
x=7

In the first printf statement x is incremented before being passed to printf so the value 6 is output, in the second x is passed to printf (so 6 is output) and then incremented and the 3rd printf statement just shows that post increment following the previous statement by outputting x again which now has the value 7.

Answer 6

i++ 使用 i 的值然后递增它，但 ++i 在使用它之前递增 i 的值。

Answer 7

The difference between post- and pre-increment is really, in many cases subtle. post incremenet, aka num++ , first creates a copy of num , returns it, and after that , increments it. Pre-increment, on the other hand, aka ++num , first evaluates, then returns the value. Most modern compilers, when seeing this in a loop, will generally optimize, mostly when post increment is used, and the returned initial value is not used. The most major difference between the 2 increments, where it is really common to make subtle bugs, is when declaring variables, with incremented values: Example below:

int num = 5;
int num2 = ++num; //Here, first num is incremented, 
                  //then made 6, and that value is stored in num2;

Another example:

int num = 5;
int num2 = num++; //Here, num is first returned, (unfortunately?), and then 
                  //incremented. This is useful for some cases.

The last thing here I want to say is BE CAREFUL WITH INCREMENTS . When declaring variables, make sure you use the right increment, or just write the whole thing out ( num2 = num + 1 , which doesn't always work, and is the equivalent of pre-increment). A lot of trouble will be saved, if you use the right increment.

Answer 8

it does not matter if you use pre or post increment in an independent statement, except for the pre-increment the effect is immediate

//an example will make it more clear:


int n=1;
printf("%d",n);
printf("%d",++n);// try changing it to n++(you'll get to know what's going on)

n++;
printf("%d",n);

output: 123