Help from a C language novice: After executing the statement for(i=0;i++<3;) printf("%d",i);, what is the program output?

The program execution process is as follows

The first time: i=0, compare 0<3, then i++, i=1, execute the output 1

The second time The third time: i=1, compare 1<3, then i++,i=2, execute output 2

The third time: i=2, compare 2<3, then i++,i=3, execute output 3

The fourth time: i=3, comparison 3<3 is false, jump out of the loop and end

Therefore, the final output is 123. Help! What is the value of variable i after executing the statement "for(i=1;i++<5;);" in c language?

The value of variable i is 6 after the c language executes the statement "for(i=1;i++<5;);"

Because when i==4, (then i becomes 5) The judgment is true and the loop is executed

When i==5, (then i changes to 6) the judgment is false and the loop exits int i; for(i=0;i++< 10;i++)i++; printf("%d",i); Why does the output i in this program be 13?

When the third loop is completed, i==9

In the fourth loop, it is judged that i++<10, which is 9<10, which is true, and then i++ is executed three times, so i=12

In the fifth loop, it is judged that i++<10; which is 12<10, which is False, then i=i+1, and finally i==13

It may be clearer if you change it to this

This is the same result

int i= 0;

while(i++<10)

{

i++;

i++;

} C Language main() {int i, a[10]; for (i=9;i>=0;i--)} a[i]=10-i; printf("%d%d%d"a[ 2],a[5],a[8],) output result

There is an error in this program. . It should be like this:

#include

void main()

{

int i , a[10] ;

for (i=9;i>=0;i--)

a[i]=10-i;

printf("% d%d%d",a[2],a[5],a[8]);

}

The output is 852 C language problem with program fragment: int i =0;while(i++<=2); printf("%d\n",i);What is the correct execution result?

When entering for the first time, i++ = 0 <= 2; At this time, i changes to 1

When entering for the second time, i++ = 1 <= 2; At this time, i changes is 2

When entering for the third time, i++ = 2 <= 2; At this time, i changes to 3

When entering for the fourth time, i++ = 3 > 2; while terminates, i changes to 4

Print, so the result is 4;

There is also a simple way; the escape condition of i++ >= 2 is i++ = 3; Obviously at this time i =3; I did ++ again; so it will

become 4, and it will be 4 naturally when printing.

c language main() { int i=3,j=0; j=(++i)*(i++)+(--i)*(i--); printf("j =%d",j); }

18 Execute the following C language program segment. How many times will the loop body be executed? for(i=0,j=0;(j!=4)‖(i＜3);j＋＋,i＋＋)printf("%d",j);

There are two conditions for looping :

j != 4 Satisfies when j = 0, 1, 2, 3, 5, ...

i < 3? Satisfies when i = 0, 1, 2 < /p>

The two conditions are in an OR relationship, and will stop when neither is satisfied.

So when j = 4, it is the only condition that neither is satisfied. So it will Execution? 0, 1, 2, 3 These four times, there is an integer variable i in the C language, and its value is 025. Why is the result 21 after executing the statement printf ("%d", i);?

At the beginning of 0, he thinks you entered the octal system.

Then you use the %d decimal placeholder to output. 25 in octal system is 21 in decimal system. =3, the output result after executing the statement printf("%d",(a=3*5,a*4),a+5) is

You will know it by executing it yourself. . .

After executing the statement "for(i=0;i++<3)", the value of variable i is

After execution, i=5i=6; because a loop will be executed after i=5 , so i++ will change i to 6; specifically, you can execute the following code to see: #includemain(){inti;for(i=2;i++<5;){}printf("%d\n",i );}