Int a [9]=,2,3,4,5,6,7,8,9 {1};
Int b [9]=,2,3,4,5,6,7,8,9 {1};
Int * p=a;
Int * p1=b;
for (int i=0; I<9. I++)
{
For (int j=0; j<=I; J++)
{
cout <* (p + j) & lt; <"*" & lt; <* (p1 + I) & lt; <"=" & lt; <* * * (p + j) (p1 + I) & lt; & lt;" \ t ";
}
Cout}

CodePudding user response:

Good, I do not know to ask shrimp, or want to share,

CodePudding user response:

reference 1/f, the wise know already should good karma response:

well, I do not know to ask shrimp, or want to share,

Can share

CodePudding user response:

Make a suggestion, now that you are a, b are the same, why not just use the same?

 
Int a [9]=,2,3,4,5,6,7,8,9 {1}; 
Int * p=a; 
Int * p=a; 

CodePudding user response:

Share it took points

CodePudding user response:

The

reference 3 floor ytfhwfnh response:

make a suggestion, now that you are a, b are the same, why not just use the same?

 
Int a [9]=,2,3,4,5,6,7,8,9 {1}; 
Int * p=a; 
Int * p=a; 

So good ah accept advice an array called twice

CodePudding user response:

reference 5 floor m0_50545500 reply:

Quote: refer to the third floor ytfhwfnh response:

make a suggestion, now that you are a, b are the same, why not just use the same?

 
Int a [9]=,2,3,4,5,6,7,8,9 {1}; 
Int * p=a; 
Int * p=a; 

So good ah accept advice an array called twice

The two words can be renamed ideographic but arrays

CodePudding user response:

An array of all don't need to use,

CodePudding user response:

It doesn't matter now hardware resources to run these small program!