Ask me about the C language! Solve C language problems!

In fact, it is much easier to read by adding parentheses to the first part of your formula! They are the same!

k=（x++>=0）&&y--<=0)||z=x+y) here the first = sign, && and ||Divide the whole equation into four pieces, &&& and ||, according to the priority of the operatorsThe operation level is the same as the order from left to right, and the && operator will be executed when the operation formula on the left is true (=1), so the operation on the right will be executed, so the operation of x++>=0 is -1 after running>=0 is false, so the sentence y--<=0 is not executed, but z=x+y=0+5=5, and finally z=0||1=1

k, x, y, z are 1, 0, 5, 5, respectively

The answer in the book is not wrong, I ran it, and the result is also 1,0,5,5 There is a misunderstanding that the boolean value of z=x+y is not 0, but 1, and the boolean value of all assignment statements is, because &&& is better than ||'s priority. So k is equal to 1

I tested it with a program and found that x++>=0 first determine whether x is greater than 0, and then ++.

I can only explain it here, and I don't quite understand why for the rest of it.

You can use the program to see why y=5,z=5

Why do you have to ask two questions for one question?

The answer is given, in:

For this kind of problem, from the syntactic analysis alone, the program execution will be different, char **pstr; is defined as a secondary pointer of type char; The assignment is conditional on the same type, or the types can be automatically converted between them.

Yes. 1) *pstr, which dissolves the first-level pointer with *, so *pstr is a pointer to the char type. i.e. there is *pstr type which is char*, and ,"hi"Its type is char and it can be converted to char *, which can be assigned.

Same as char *p, a[5]; p=a; Grammatically it is correct.

2) pstr is a secondary pointer, and its type is char **hi"Its type is char and it can be converted to char *, but obviously char ** and char* cannot be converted to each other, so it is actually wrong.

3) **pstr, which dissolves the pointer with **, it is of type char, and 2) in the same way, char and char or char * are obviously not convertible to each other, so they are actually wrong.

printf("%d",x-

;Here is the value of the expression that outputs x--, but here because it is a self-decreasing expression, and x is on the left, the value of the expression is still the value of x, so the output statement is executed, and the output result is 3, and then executed.

while(!x);Because this is the inverse, the result is 0, that is, false, so the program is executed, of course, the premise is that you ubt

x=3;This should be the int type, right?