{
Char x [80]="AB", [80] y="LMNP";
Int n=0;
Strcat (x, y);
While (x [n++]!='\ 0')
[n] y=x [n].
Puts (y); }
A, LBB, ABLMNPC, ABD, LBLMNP
To ask, the answer is LBLMNP, but I can't understand the first L come from,

CodePudding user response:

While (n++) this sentence, judge 0, then become 1 + +, after entering the first sentence is [1] y=x [n], so y [0] remain the same

CodePudding user response:

The problem here

 while (x [n++]!='\ 0') 

Cycle for the first time, n=0, since is a suffix, so the judgment on whether the x [0] is equal to '\ 0', the n value into 1, then execute the while loop body at this time, n=1

 [n] y=x [n]. 

that is [1] y=x [1], [0] y previous value has not been x [0] cover ,
Change the code to the

 while (x [n].='\ 0') 
{
[n] y=x [n]. 
n++; 
} 

Can realize to complete modification of y,

CodePudding user response:

refer to the second floor MianHou response:

problem here

 while (x [n++]!='\ 0') 

Cycle for the first time, n=0, since is a suffix, so the judgment on whether the x [0] is equal to '\ 0', the n value into 1, then execute the while loop body at this time, n=1

 [n] y=x [n]. 

that is [1] y=x [1], [0] y previous value has not been x [0] cover ,
Change the code to the

 while (x [n].='\ 0') 
{
[n] y=x [n]. 
n++; 
} 

Can realize to complete modification of y,

Thank you for your duck duck

CodePudding user response:

reference 1st floor ggglivw response:

while (n++) this sentence, judge 0, then become 1 + +, after entering the first sentence is [1] y=x [n], so y [0] intact

Thank you do ~