C language, which I can t understand, do me a favor

10%10 = 0 Why is it equal to 0 Because % is the remainder of 10 divided by 10 and the remainder is 0.

c stands for what does it mean, and what does it mean, and why is the value of + 4 equal to 5?

c is the meaning of c=c+1. ++ means plus 1. ++4=4+1。

But note a little ++c at 21 and c++ at 20. But the c's are all 21. The difference is between ++ and last, if you calculate the value of adding 1 first, and take the value of c to represent this equation first.

Distinguish between ++c and c++. If c=20;Then do these two equations, then yes.

C is 21 for the result, C is 20 for the result, and C is 21.

10%10 = 0 means finding the remainder of 10 divided by 10 It must be 0+c means that the variable c adds 1, ++ is the variable self-addition operation, ++4 is 4+1 of course 5.

A variable is a number stored in a computer, represented by one or several letters, and its value is variable) a=3; b=5;

c =++ a* b;Equivalent to a=a+1; c=a*b;

Note that now a=4 b=5 c=20).

d =a + b;Equivalent to d=a*b;

At this point, a=4, b=5, d=20).

a=a+1;

At this point, a=5, b=5, d=20).

This is the modulo division operator.

Modulo division is the division of two integers to take the remainder; It is required that both sides of % must be integers;

The difference between a and a++.

a is equivalent to a=a+1;

a++ is equivalent to a=a; Then a plus 1

10%10 means the remainder of 10 divided by 10, which is of course 0.

is self-adding, ++c is adding 1 to c before the operation, and c++ is adding 1 after the operation

Hehe, I'm also learning C++ for a month, and I'm also a beginner.

i = i %= j has a value of 0i = 3 2 * j = 3 - 2).

The same question answers.

Question 1 Analysis:Variables: int i = 10, j = 5;

The priority of operators = and %= is the same level, and the combination is from right to left, so first calculate the %= operator on the right, that is, i%=j first, this expression is equivalent to i=i%j, i = 10 % 5, the result is 0, and then 0 is assigned to variable i, so variable i

The value of 0 is . Analysis of the second question:Calculate the parenthetical expression first, then j = 1, then 3 2 * 1, because the result of 3 2 is 1 (both are integers), so the value of variable i is 1.

Test**.

Need to understand here"="The number is an assignment number, not an equal sign, in the C language"=="is equal to. and take the remainder in the formula, int i=10, j=5;

So i=i j, i=10 5=0, i=3 2*(j=3-2), you need to calculate the right side first, and assign the value to i, i.e., i=3 2*1 (in mathematics, i is ultimately equal to, but because i is an integer, i=1 is said

This is a complex pointer with a right-to-left reading rule. Reading: Start with the identifier (as in the t, read to the right, encounter the back parentheses) and then to the left. From the inside out, until the end of the reading.

For example, t, 1) right reading, t is an array of 10 elements. In case of inverted parentheses, read to the left, the element type of the array is the pointer.

2) Where does the pointer point? Reading to the right, I came across [10], which turned out to point to an array of 10 elements.

3) The element type of this array is int*

However, it is obvious that the meaning of the title does not need to be read at all, because it is enough to find the exact same form, and it is obviously d.

Complex pointers are generally used in DLLs, low-level calls, etc., and are not usually used.

It doesn't make much sense even if you figure it out, who defines types like that, if I give up.

The meaning of this problem is to count the number of numbers from 0 to 9 in the string str, where the ten elements of array a store the numbers from 0 to 9.

The difficulty is in A[(*P-48)]. 48 is the ASCII code of '0', '1' is 49, and so on.

p points to each character in the str. When *p is one of 0-9, e.g. *p='1'then *p-48='1'-48=49-48=1。Then a[1]++ will be executed, i.e. the number of occurrences of '1' will be +1

Answer: 0-0 1-4 2-2 3-2 4-2 5-0 6-2 7-2 8-0 9-0 is to count the numbers from 0 to 9 in the string above. For example, 1-4 means that there are 4 1s

Hello. This is a more basic C language practice question. Of the three loops, the first loop is array initialization, i.e. all values are assigned 0.

The core function lies in the second loop, which is function 1Scan the entire str array, 2If the ASCII code of a character is 0-9, the value of the array corresponding to that value is +1.

That is, the function is to count how many times each number appears.

The final output format will be: 0-0 1-4 2-2,。。 It means that the number 0 appears 0 times, the number 1 appears 4 times, and the number 2 appears 2 times.

I hope you will be patient and careful to understand the procedure according to my description, and I hope you can understand him.

Count the number of occurrences of each number from 0 to 9.

C language: General Programming Language.

This is a computer programming logic language, you need a system of learning, if you really want to understand, you can buy some C language books to learn, there are also courses on the Internet, but if you are not bad money, you can sign up for a class, this is the best. Thank you.

1）b='a'Equivalent to b = 97 ('a'The ASCII code is 97) C>3 is true, and then the value after && is calculated.

A value of 0 is 1 (0 is false, non-"false" is "true", and a value of 1) 8<1 is false.

So the first value is 0

2) A value of f 3 that is not 0 is true).

The value of a-b is -95

The second one has a value of 1

3）!(95) has a value of 0

4) a=7 is an assignment expression with a value of 7

a=7)>6 (true).

b=-1) has a value of -1

b=-1) >6 is false, and the fourth has a value of 0

One of the computer programming is not easy to learn when I was in school, but now I can't use it, so I still have to rely on my own hard work.

The following variable declaration has static, indicating that these two variables will not be redefined until the end of the program, so what is the value of the variable after the first call, and the second continuation of the value of the variable after the first call.

After the first execution, m=

The second time, i = 3 + 8 + 1 = 12 and m = 12 + 4 + 1 = 17