What is the result of this C language question, ask for an explanation

The program is wrong, and the result after running is 5, 5, 5, 5 .........Infinite 5) The following places are wrong:

# include

# define n 8

void fun(int * x,int i)* x=*(x+i);}

main()

int a[n]=，i；

fun(a,2);

for(i=0;i=n/2;i++) shilling i=0, the second one should be a bool expression, you wrote an assignment expression, and there is an error here.

printf("%d",a[i]);

printf("");

fun(a,2) is to pass in the address of a[0], and the function does a[0]=a[2].

Then output half of the array, 3,2,3,4

# include

# define n 8

void fun(int * x,int i)* x=*(x+i);}

main()

int a[n]=,i;

fun(a,2);After running: the result of a is: a[0--7]= that is, a[2]--a[0].

for(i=0;i=n/2;i++) It's an endless loop here. Is it to iprintf("%d",a[i]);

printf("");

The results are all 5

There's a problem!

Here's ** to write it more clearly:

if(a>b)

c=a;a=b;

b=c;If only one sentence is included in the compound statement: c=a; The judgment of A>B is obviously not valid, and this sentence will not be enforced.

After that, the remaining two sentences are executed sequentially:

a=b;A is 10

b=c;b is 16

So the end result:

a = 10, b = 16, c = 16 if changed to:

if(a>b)

Here is that the if compound statement contains 3 sentences, the 3 sentences are not executed if the conditions are not established, and the final result of the 3 variables is unchanged.

Because a b is false, c=a is not executed

So execute a=b, so a=10

Then b=c, so b=16

So in the end it's 10, 16, 16

There are no special curly braces, so only the last two sentences are executed.

i.e. a=b; b=c;

Note that it's an assignment, not an equal.

Then the final answer is d

If the statement does not have a curly brace enclosed number, only the first " is executed;

If a>b, c=a; And a just executes a=b; b=c;

a=10;b=16;

c=16

Here the if statement does not add curly braces, so the conditional judgment of the if statement only corresponds to c=a; This sentence.

Because a only executes a=b; b=c;

So the answer is d

If not true, the first step is not involved, operation 2 3 steps are enough, 2: a=b a=10 b=103: b=c b=16 c=16 finally a=10 b=16 c=16

If the poster is sure that the program is okay, the answer is: d

Calculate and output numbers between 100 and 150 that are divisible by 3, in a row of 10.

This is the relationship between the structure and the common body, basically this structure is that each byte has its own internal error space, and the common body is all the functions, a piece of no space, you now assign a value of 1234, that is, the common body is 1234, because the whole common body is in the memory applied for by word, that is, 4 bytes, the first two bytes of the four bytes are occupied by the structure, that is to say, low and high each occupy one, and the printing result is: word= 1234, high=03, low=04, the last one is 12ff Hope it helps.

S is reassigned and accumulated each time in the loop body, so only the last round of loops is calculated.

k=5,s=1+5=6

Select Answer B