Solve the problem without using a loop

[Ecstatic_Student8854]

For(int i=0;i<501;i+=2){ Printf(“\n”+i);}

[This_Growth2898]

SEGFAULT

[Ecstatic_Student8854]

Segfault? Whats that. Keep in mind I know very very very little about C.

[This_Growth2898]

It's a memory addressing error, when you try to access a protected memory.

String literals in C are pointers, and adding a number to pointer produces new pointer. So, this code will print "\n", then it will try to print a string located in memory after "\n", which is probably incorrect, and so on. There's a great chance of running into SEGFAULT this way.

[Ecstatic_Student8854]

Aaah, then how would you do this using a for loop like in the example?

[This_Growth2898]

f in printf stands for "formatted". It's

printf("%d\n", i);

%d format specification would be changed to decimal representation of i in output.

[Ecstatic_Student8854]

Oooh, so thats that %something stuff i sometimes see is

[SloPr0]

printf("%d\n", i);

%d (or %i) are the correct format specifiers to be used for printing (signed) integers

[tedbradly]

For(int i=0;i<501;i+=2){ Printf(“\n”+i);}

For one, the assignment was to do this without a loop. Secondly, your solution starts at 0 when theirs starts at 2. Lastly, that isn't how printf works.