Is another for loop allowed in the counter section (third part) of a for loop? In my attempt to get elegant in writing code to produce a right triangle, I wrote this but it wouldn't compile:
#include <stdio.h>
int main()
{
int i, j, N = 5;
for (i = 1;
i <= N;
(for (j = 1; j <= i; j++, printf("%c", '0'));), i++)
printf("\n");
}
return 0;
}
No there are allowed only expressions or declarations.
EDIT: I am sorry. I thought you are speaking about the condition part of the loop. In the expression part of the loop there are allowed only expressions.
You could use a lambda expression that would contain this for loop. For example
for ( i = 1;
i <= N;
[]( int i ) { for ( j = 1; j <= i; j++, printf("%c", '0' ) ); }( i ), i++)
Here is a demonstrative example
#include <iostream>
int main()
{
int N = 10;
for ( int i = 1;
i < N;
[]( int i )
{
for ( int j = 1; j < i; j++, ( std::cout << '*' ) );
}( i ), i++ )
{
std::cout << std::endl;
}
return 0;
}
The output is
*
**
***
****
*****
******
*******
********
Or your could define the lambda expression outside the outer loop that to make the program more readable. For example
#include <iostream>
int main()
{
int N = 10;
auto inner_loop = []( int i )
{
for ( int j = 1; j < i; j++, ( std::cout << '*' ) );
};
for ( int i = 1; i < N; inner_loop( i ), i++ )
{
std::cout << std::endl;
}
return 0;
}
Take into account that in general case the nested loops showed in other posts are unable to substitute the loop with the lambda-expression. For example the outer loop can contain continue statements that will skip the inner loop. So if you need that the inner loop will be executed in any case independing on the continue statements then this construction with the lambda expression will be helpful.:)
There is no need to do so. Because for loop can be easily replaced with while loop, every part of for loop can be placed in another place, where it's possible to use complex constructions. In your case, you can just change loop to the following:
for (i = 1; i <= N; i++) {
printf("\n");
for (j = 1; j <= i; j++) {
printf("%c", '0');
}
}
However, if you really have to place complex action, you may use gcc extension (compound statement):
for (i = 1;
i <= N;
({for (j = 1; j <= i; j++) putchar('0'); }), i++) {
printf("\n");
}
lambdas as the other answers do.
for is executed after each loop. That's why printf() is not called last time.In the counter section of a for() loop, expressions are allowed, but statements are not.
And every for() line in C/C++ forms a new statement (it's not an expression).
However, you can nest several for() loops if you want.
For example, since you want a new loop in the counter sections, that means that you need to perform a loop at the end of the main for() loop.
This is the scheme:
for (int i = 0; i < i_max; i++) {
// stuff...
for (int j = 0; j < j_max; j++) {
// stuff..
}
}
You can't do that because condition and increment parts of a for can only contain expressions. A for loop is an iteration statement, though.
Simply nest the loops like sane programers do:
#include <stdio.h>
int main()
{
int N = 5;
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= i; j++)
printf("0");
printf("\n");
}
}
If you're not feeling well, though, you could use a lambda:
#include <stdio.h>
int main()
{
int N = 5;
for (
int i = 1;
i <= N;
[=](){ for (int j = 1; j <= i; j++) printf("0"); }(), printf("\n"), i++
) ;
}
Elegance comes with clarity.
When I want to create a string of characters, I construct a C++ object called std::string.
#include <iostream>
#include <string>
int main()
{
char c = '0';
const int n = 5;
for (int i = 1; i <= n; ++i)
{
std::cout << std::string(i, c) << '\n';
}
}
So there is no need for a nested for-loop in this particular case.
Otherwise put a for-statement in the body of the outer loop as other answers suggested.
Shortest-possible solution:
main(i){for(i=1;i<11;printf("%0*d\n",i++,0));}
Output:
0
00
000
0000
00000
000000
0000000
00000000
000000000
0000000000
\n into the printf statement, and make it even smaller.
Hat tip to Michael Burr for the suggestion to use lambda. And thanks to the commentators requesting me to use putchar().
#include <stdio.h>
int main() {
int N;
scanf("%d", &N);
for (int i = 0; i < N; i++, [i] {
for (int j = 1; j <= i; j++, putchar('0'))
;
}(), printf("\n"))
;
return 0;
}
I think we should avoid obfuscated code unless we get paid for making the code complex.
i,j, andN? 2. What is the value ofN? 3. Where does the 3rd side get calculated?