C-Language arithmetic operators

The basic arithmetic operators for the C language are the following table:

name	symbols	Description
Addition operator	+	Binocular operators, that is, should have two of the amount involved in the addition operation. such as a+b,4+8 and so on. has right-associative nature.
Subtraction operator	-	Binocular operator. But "-" can also be used as a negative operator, which is a single-mesh operation, such as-x,-5, such as a left-associative.
Multiplication operator	*	Binocular operator with left-associative.
Division operator	/	Binocular operator with left-associative. When the participating operands are integer type, the result is integer type, and the decimal is rounded off. If one of the operands is a real type, the result is a double-precision solid.
Remainder operator (modulo operator)	%	Binocular operator with left-associative. The amount required to participate in the operation is integer and cannot be applied to a float or double type. The result of the remainder operation is equal to the remainder after the division of two numbers, and the result is 0 when divisible.

Binocular operators + and-have the same precedence, their precedence is lower than the precedence of operators *,/and%, while the precedence of operators *,/and% is lower than the precedence of the single-mesh operator + (plus sign) and-(minus).

1#include <stdio.h>2 intMainvoid){3printf"\n\n%d,%d\n", -/7,- -/7);4printf"%f,%f\n",20.0/7,-20.0/7);5     return 0;6}

In this example, the results of 20/7,-20/7 are integer and all decimals are removed. and 20.0/7 and -20.0/7 because of the real number to participate in the operation, so the result is also real type.

#include <stdio.h>int main (void) {    printf ("%d\n", 3%);     return 0 ;}

This example outputs 100 divided by 3 for the remainder 1.

Self-increment, decrement operator

The self-increment 1 operator is recorded as "+ +", whose function is to increment the value of the variable by 1, and the decrement 1 operator as "--", whose function is to reduce the value of the variable from 1.

Since the increment of 1, the self-minus 1 operator is a single-mesh operation with right-binding. These can be in the following ways:

• The ++i:i is added 1 and then participates in other operations.
• --I:I 1 and then participate in other operations.
• After the I++:I participates in the operation, the value of I is increased by 1.
• After the I--:I participates in the operation, the value of I is reduced by 1.

Error-prone in understanding and use is i++ and I--。 Especially when they are in more complex expressions or statements, they are often difficult to figure out and should therefore be analyzed carefully.

#include <stdio.h>intMainvoid){    intI=8; printf ("%d\n",++i); printf ("%d\n",--i); printf ("%d\n", i++); printf ("%d\n", i--); printf ("%d\n",-i++); printf ("%d\n",-i--); return 0;}

The initial value of I is 8, the 4th row I plus 1 after the output is 9; the output of the 5th line minus 1 is 8, the 6th line outputs I is 8 followed by 1 (9), the 7th line output I is 9 minus 1 (8), the 8th line output 8 is 1 (9), and the 9th line output 9 minus 1 (8).

1#include <stdio.h>2 intMainvoid){3     intI=5, j=5, P,q;4P= (i++) + (i++) + (i++);5Q= (++J) + (++J) + (+ +)j);6printf"p=%d,q=%d,i=%d,j=%d", p,q,i,j);7     return 0;8}

In this program, the p= (i++) + (i++) + (i++) should be understood as three I add, so the P value is 15. I then increment 13 times to add 3 so the last value of I is 8. For the value of Q is not, q= (++j) + (++J) + (++J) should be understood as J to increment 1, and then participate in the operation, because J since 13 times after the value of 8, three 8 added and the last value of 24,j is still 8.

For p= (i++) + (i++) + (i++), the first calculation (i++) + (i++), because it is "post-add", equivalent to 5+5, the result is 10, and then calculate the + + (i++), equivalent to 10+5, the result is 15.
For q= (++J) + (++J) + (++J), first calculate (++j) + (++J), because is "pre-add", to calculate two times ++j, at this time j=7, then add, equivalent to 7+7, the result is 14; then the 14+ (++J), the equivalent of 14+8, the result is 22.

C-Language arithmetic operators