Explore the c language in front ++ and back ++, explore c

Explore the c language in front ++ and back ++, explore c

Wavelet takes you to explore the front ++ and back ++ in the C language. You are welcome to vomit, and you are welcome to add QQ463431476. Welcome! Now let's explore: Let's first check that the first I is assigned 0, And I ++ (after ++) does not output 1. Now I is assigned 0, ++ I, that is, the first ++ outputs 1. I is assigned a value of 0. The first ++ is to add 1 first and then assign a value to 1. The values of the following I are all 1. I is assigned 0, followed by ++, followed by ++, which means that I am assigned a value first. After the value is assigned, I will add 1 after the end of this statement. The first printf outputs the value 0 from which I starts to be assigned. The first printf is followed by a semicolon (;), which indicates the end of the statement. When this statement ends, I will secretly add 1, and the 2nd printf outputs the I value after the previous statement is processed. I is assigned 0, I ++; this is a statement. Because there is a semicolon. After adding a value, we assign a value first, and then add another value when the statement ends. I ++; the execution result is 0 + 1 = 1. Printf only outputs the I value. I is assigned 0 first, And I ++ is followed by a semicolon ";", that is, this statement ends, I ++; the result is 0 + 1 = the value of I after 1 is 1. I is assigned 0, ++ I. The first ++ is to add 1 first, and then assign it to itself. ++ I; the result is 0 + 1 = 1. I first assigns a value of 0, ++ I; (first ++) means that I first 0 + 1 = 1, and then assigns a value of 1 to myself. I becomes 1 I ++; (After ++) I has been assigned 1. Here I ++; that is, first assign the value 1 of the previous I to myself as 1, however, a semicolon (;) indicates the end of this statement. When the statement ends with I ++;, 1 + 1 = 2 is executed. I is assigned 0, front ++, I will be 1. The I ++ output in printf can only be the value assigned by itself. 1. If you add another printf ("% d", I) below ); the output I is 2. To sum up, add 1 to the front ++ and assign the value to the front ++. + + Is to assign values by yourself. After the value is assigned, you must wait until the statement ends before adding 1. Tip: prefix + Postfix. This is what I think. It means that the first step of ++ is plus + 1, and then the last plus + + is assigned a value first. In addition, the principle of prefix and suffix is the same.