CodePudding user response:

 # include & lt; Stdio. H> 
Int main () 
{
Int I, j, k; 
for(i=1; I<=4; I++) 
{
for(j=0; J{
Printf (" "); 
} 
//for (k=0; k<5 - I; K++)//this is k=0, perform k<5 - I printf (" * "), k++; 
//I=1, k=0 & lt; 5 - I==4, 5-1 condition is met, execute the printf (" * "), k++; 
//I=2, k=0 & lt; 5 - I=5-2=3, the condition is met, execute the printf (" * "), k++; 
//I=3,,,,,,,,, 

For (k=5 - I; K> 0; K -)//should it in reverse, k=5 - I, perform 5 - i> Zero printf (" * "), k -; 

//for (k=4; k<5 - I; K -)//why the above for (k=0; k<5 - I; K++) with my comment is not ah, solving? 
{//when I=1, the condition is k=4, k=4 & lt; 5 - I==4, 5-1 conditions are not met, just jump out of the loop the 
Printf (" * ");//when I=2 k=4 & lt; 5 - I=5-2, still does not meet the conditions, jump out of the loop 
}//I=3,,,,,,,,, 
Printf (" \ n "); 
} 
system("pause"); 
return 0; 
}