C language look at the program problem, C language read program problem?

In the C language. An integer constant starting with 0 refers to octal. 027 is 2x8+7, so the answer is c.

Question 5, m=027 where 0 represents octal, and the output is changed to decimal, i.e. 2 8+7=23.

Sixth, since the statement belongs to the logical judgment that the result is only 0 or 1, and because the values that are not 0 are all true, that is, 1. So it can be seen as 1&&1 1=1

Question 7, x+=x%=(6)+4 can be regarded as:

x%=-2,x+=x

According to the above order, the first x = 10% (-2) = 0 This is x changed from 10 to 0, so the second x is equivalent to x = 0 + 0 = 0 Read more books by yourself, and it is not difficult to think more.

First of all, you need to know a few things:

1. printf("%d",--x) and printf("%d", x--)

The former calculates x=x-1 and then outputs x, while the latter outputs x=x-1 and then calculates x=x-1

2.The role of continue:

continue is to end the loop and skip the unexecuted statement below the loop body.

In this case, you are skipping printf("%d,",--x);This line goes back to for(;; x>0 ;x--) line.

3.if(0) and if(non-zero):

A value of 0 is false, and a value that is not zero is true. Therefore, the statement in if(0) is not executed, and the statement in if(non-zero) is executed.

process (each step is every time to determine whether x>0 is true or not):

1.At this time, x=8, because x%3=non-zero, so the output x value is 8, and then x=x-1 is calculated, and continue returns to for(; x>0;x--) This line calculates x--

2.In this case, x=6, because x%3=0, printf( is executed"%d,",--x);This line calculates x=x-1 and outputs the x value of 5

Finally back to for(; x>0;x--) This line calculates x--

3.In this case, x=4, because x%3=non-zero, so the output x value is 4, and then x=x-1 is calculated, and continue returns to for(; x>0;x--) This line calculates x--

4.At this time, x=2, because x%3=non-zero, so the output x value 2, and then calculate x=x-1, continue back to for(; x>0;x--) This line calculates x--

5.At this time, x=0, at this time, x>0 is no longer established, and the program is over.

x=8 x>0 8%3 is 2 if is true [output x-- is 8 ] x is 7 after executing continue x-- x is 6

x=6 x>0 6%3 is 0 if is not true [output --x is 5] x is 5 execute x-- x is 4

x=4 x>0 Hold 4%3 is 1 If Hold [Output x--is 4] x is 3 Execute x-- x is 2

x=2 x>0 Hold 2%3 Hold 2 If Hold [Output x--2]x is 1 Execute x-- x is 0

x=0 x>0 does not hold ends.

So the output is 8 5 4 2

First x = 8

for loop.

First time x = 8

x%3 = 2 if statement is true.

Enter x-- output 8 first, then x-- x=7, second x-- after x= 6

x%3 = 0 if statement is not true.

Execute the output statement --x x=5 and then output 5

After the third x--after x= 4

x%3 = 1 if statement is true.

Output 4 first and then x-- x=3

After the fourth x-x=2

The 2%3 = 2 if statement is true.

Output 2 first and then x-- x=1

After the fifth x-- x = 0

x>0 does not hold to exit the loop.

End of program.

The first time x=8, x%3=2 is true, and the expression is 8 after x--, 8 is printed, but x becomes 7, and the following printf does not need to be executed.

to the for loop x--, which becomes 6 x%3=0 is false Execute printf("%d,",x); The expression for x is 5

After going to the for loop x--, it becomes 4 4% 3=1 is true Execute the if statement Print printf("%d,", x--) is 4 and x is 3

to the for loop x--, which becomes 2 2% 3=2 is true Execute the if statement Print printf("%d,", x--) is 2 and x is 1

After the for loop x---, x=0 does not meet the condition and jumps out.

Determine whether x>0 is true, and terminate the function if x>0 is not true. If x>0 is true, then determine whether x%3 is 1, if so, execute x--x-in the for statement--if x%3 is not 1, execute printf("%d,",x);