Of the new turn-CodePudding

C, if has the following definition and assignment statements
Int s [2] [3]={0}, (* p) [3].
P=s;
The p + 1
* (p + 1)

CodePudding user response:

P + 1 times line first address line (i.e., s [1])
* (p + 1) s [1] [0] element address

Both the value of the goods is consistent, different meanings,

CodePudding user response:

This code is run, you'll soon see,

 int main (int arg c, char * argv []) 
{
Int b [3] [4]={{1 hc-positie}, 
9,11,13,15 {}, 
17,19,21,23 {} 
}; 

Printf (" 1 b [2] [1]=% d \ n \ n ", b [2] [1]). 
Printf (" [1] 2 b=% d, % d \ n \ n ", [1], b * b [1]). 
Printf (" 3 b=% d, % d \ n \ n ", b, * * b); 
Printf (" 4 + 1 b=% d, % d \ n \ n ", b + 1, * * (b + 1)); 
Printf (" 5 * b + 1=% d, % d \ n \ n ", * b + 1, * (* b + 1)); 
Printf (" 6 * (b + 1)=% d, % d \ n \ n ", * (b + 1), * (* (b + 1))); 
Printf (" [2] 7 b + 1=% d, % d \ n \ n ", [2] + 1, b * [2] (b + 1)); 
Printf (" 8 * (b + 1) + 1=% d, % d \ n \ n ", * (b + 1) + 1, * (* (b + 1) + 1)); 
Printf (" 9 * (* (b + 2) + 2)=% d, % d \ n \ n ", * (b + 2) + 2, * * (b + 2) + (2)); 
system("pause"); 
return 0; 
}

CodePudding user response:

Modified the print line 9:

 printf (" 9 * (b + 2) + 2=% d, % d \ n \ n ", * (b + 2) + 2, * * (b + 2) + (2));