C program interpretation, C language program interpretation

char *p;

uchar log_type;

ulong querytime, log_total, count;

uint wlen, rlen;

char wbuf[wbuf_size], result[rbuf_size];

dummy_t root;

dstuos_node *ptr;

Define global variables).

log_type = 1; start_rows =0; end_rows = page_size - 1; count = 0;

Assign an initial value to the variable).

else if (ret == 2)

else if (ret == 3) continue;

else return -1;

judgment variable ret).

The program you gave is incomplete, and the specific functions are not clear, as if it is a server program.

You're not working on a graduation project, are you?

t=t*p%c;The % sign has the highest priority, and t = t * 2 when i = 1 %2009 = 1*2 =2

s=(s+t)%c;(Highest priority.)

And so on. So the result is as follows:

i=1 t=2

j=1 s=2

i=2 t=4j=1 s=6 j=2 s=10

i=3 t=8j=1 s=18 j=2 s=26 j=3 s=34

i=4 t=16j=1 s=50 j=2 s=66 j=3 s=82 j=4 s=98

i=5 t=32j=1 s=130 j=2 s=162 j=3 s=194 j=4 s=226 j=5 s=258

i=6 t=64j=1 s=322 j=2 s=386 j=3 s=450 j=4 s=514 j=5 s=578 j=6 s=642

i=7 t=128

j=1 s=770 j=2 s=898 j=3 s=1026 j=4 s=1154 j=5 s=1282 j=6 s=1410 j=7

s=1538

i=8 t=256

j=1 s=1794 j=2 s=41 j=3 s=297 j=4 s=553 j=5 s=809 j=6 s=1065 j=7 s=1

321 j=8 s=1577

i=9 t=512

j=1 s=80 j=2 s=592 j=3 s=1104 j=4 s=1616 j=5 s=119 j=6 s=631 j=7 s=1

143 j=8 s=1655 j=9 s=158

i=10 t=1024

j=1 s=1182 j=2 s=197 j=3 s=1221 j=4 s=236 j=5 s=1260 j=6 s=275 j=7 s

1299 j=8 s=314 j=9 s=1338 j=10 s=353

i=11 t=39

j=1 s=392 j=2 s=431 j=3 s=470 j=4 s=509 j=5 s=548 j=6 s=587 j=7 s=62

6 j=8 s=665 j=9 s=704 j=10 s=743 j=11 s=782 782

#include""printf's header void func1(int i); Advance declaration void func2(int i) for function 1; Advance declaration char st= for function 2"hello,friend";Define an array of characters of length 12.

All arrays in C++ are counted from 0 void func1(int i).

void func2(int i)

main()

Program flow: call func1(0:) in the main function

Output st[0]->h

i<3? (i=0) -i<3

i+2 i=2 calls func2(2):

Output st[2]->l

i<3? (i=2) -i<3

i+2 i=4 calls func1(4):

output st[4]->o

i<3? (i=4) -i>3

Return! Return! Return!

Return! Return!

Return! Return!

Output line wraps. End of procedure*

Start from main(), call fun1(0) to output h first, then i to 2, then call fun2(2) in fun2, output l, then i to 4, then call fun1(4), output o, and return directly.

The whole output is: hlo

Start with the main function main, the initial value of i is 0, call func1, and pass the value of i 0 as a parameter.

In func1, first print st[i], that is, st[0], that is, the first character h of the string, and go down to the if conditional sentence, and if the current i is less than 3, then execute i+=2, which is equivalent to i=i+2 => i=0+2 => i=2, and then call the func2 function, where i is 2, and pass it in as a parameter.

In func2, first print st[2], that is, the third character l in the string, then determine the if condition, the current i is 2 less than 3 to meet the condition, execute i+=2 to get i is 4, and then call func1 and pass 4 as a parameter.

In func1, first print st[i], that is, st[4], that is, the fifth character o of the string, and determine the if condition, and find that i greater than 3 does not meet the condition, and the following statement is no longer executed.

Back in the main function main, the program ends after printing the line break.

The final result is HLO

First of all, today into func1()i is 0 output h, then into func2()i is 2, output l, and finally into func1()i is 4, output o