Knowledge points must be mastered in C Language

Knowledge points must be mastered in C Language

My personal summary: learn c quickly

1 .... ++ A and a ++:

++ A adds a ++ value first. the value assigned to variable B by adding the latter is a rather than a + 1.
2 .... When the bitwise and the same 1 are 1, the other is 0, for example: 1110 & 1010 = 1010, the bitwise or the same 0 is 0, and the other is 1, for example: 1110 | 1010 = 1110
3 .... ~ : Bitwise inversion of non-calculation operations, such :~ 1111 = 0000
4 .... For example, 1010 ^ 0110 = 110 to move one digit to the left is equivalent to multiplying 2 5 .... When the condition is a range (expression and, or expression) scanf, the variable is added with the & Symbol inside the switch as the expression, and the case is followed by a colon; when the break ends the continue, only the loop is ended.
6 .... Conditional operator? A> B? A: B is true. The former is false.
7 .... Perform the following steps to calculate the maximum common divisor of two numbers and the minimum common multiple using the Euclidean Algorithm (moving phase division:

First, divide a large number by a small number to obtain the first remainder. Then, divide a small number by the first remainder to obtain the second remainder; divide the first remainder by the second remainder and obtain the third remainder. In this way, the last remainder is used successively until the remainder is 0. In this case, the last divisor is the maximum common divisor (if the last divisor is 1, the original two divisor is the mutual prime number ).

For example, for the maximum number of 1515 and 600, the first time: 1515 divided by 600, more than 2 315; the second time: 600 divided by 315, more than 1 285; the third time: divide by 315 by 285, And the quotient 1 is more than 30. The fourth time is 285 divided by 30, and the Merchant 9 is more than 15. The fifth time is 30 divided by 15, and the Merchant 2 is more than 0. The maximum number of public appointments between 1515 and 600 is 15.

Minimum Public multiples of two positive integers = product of two integers limit the maximum public approx. Of two integers

8... Arr [I] = 10 + rand () % 90 indicates that a random number ranging from 10 to 100 is generated.
9 .... The so-called "Daffodils" refers to a three-digit number, each digit of which is equal to or equal
Itself. For example, 153 is a "Daffodils" because 153 = the power of 1 + the power of 5 + the power of 3.
10... For (; arr [I];) when the string does not know the length
11 .... Defines an array of 20 elements, which must be in the range of x = 30 -- y = 70 (with a random number), such as: loop body: a [I] = arc4random () % 41 + 30 ----------------> arc4random () % (y-x + 1) + x
12... Bubble Sorting // compare n elements with n-1 vcvc8gic8vxl + forward = "O', 'K'} char str [] =" OK "differences:
The second implies an ending character "\ 0'
14. // the easiest way to initialize character arrays is to assign characters to each element in the array one by one.

Char str [10] = {'I', '', 'A', 'M', 'h', 'A', 'P', 'P ', 'y '};

15...

// Keep taking the entered characters as long as you do not encounter carriage return

Char a = 0;

While (a = getchar ())! = '\ N '){

Printf ("a = % c \ n", );

}