sqrt table + command line argument

[Kynsin]

I need to create a program that provides a table of numbers and their square roots using a command line argument. I have the command line argument down, im just unsure how to do the square root table

[E3h7X9L8]

use setw() stream manipulator to set widgth between numbers and their square , setw is a parameterized stream manipulator and you must include <iomanip> header

here is an example so you can understand how it works:

std::cout << 5 << std::setw(15) << 5*5 ;

this would display:

5--------------25

(x13 '-' which are white spaces, i cant represent them in this post, two of them are "eaten" by '25' because the numbers or characters are written from right to left in output )

[Kynsin]

Say i was to type something like a.out -15 30 5
I need to produce a table of numbers and their square roots which begins at -15 and goes in steps of 5 up to 30. I'm still quite new and I don't really know what you mean by your statement. I always see people on this forum type cout, but my teacher tells us to use printf. I don't know how to implement the numbers from the command line such as -15, 30, 5 into a for loop or whatever i need to do

[Zhuge]

Are you writing in C or C++? If you are writing C, you use printf(). In C++, we use std::cout.

Could you post what you have so we can give tips about where to find what you're looking for?

[E3h7X9L8]

i dont really know how command line argument works .but im still gonna give you the loop

for(int i = -15; i <= 30; i += 5)
    printf("Number and square: %10d %10d \n", i, i*i);

just replace -15 30 5 with the variables you input

[Kynsin]

#include <stdio.h>
#include <math.h>

int main(int argc, char *argv[4]){
   int beg, end, step, i;
   if (argc==4) {
      sscanf(argv[1], "%d", &beg);
      sscanf(argv[2], "%d", &end);
      sscanf(argv[3], "%d", &step);
   }
   if (argc!=4) {
      printf("Error, What is your starting number?\n");
      scanf("%d", &beg);
      printf("What is your number to stop at?\n");
      scanf("%d", &end);
      printf("What is your step size?\n");
      scanf("%d", &step);
   }
   for (i = beg; i = end; i +=5) {
      printf("%d   %.2f\n", i, i*i); 
   }
}

Kind of implemented what leryss just said but yeah..

[Kevin C]

He said square roots, not squares, so:

printf("%d %.2f\n", i, sqrt(i*1.0));

[Kevin C]

Also:
for (i = beg; i <= end; i +=step)

[Kynsin]

alright, so im getting nan error which is "not a number" apparently. for exmaple sqrt of -25 would just be 5i. how would i go about doing that

[Zhuge]

You could try using the complex number functions/types the library provides, but it is going to be more complex than sticking to the reals.
http://en.cppreference.com/w/c/numeric/complex

[Kynsin]

   if ((beg < end) && (end > 0)) {
      for (i = beg; i <= end; i +=step) {
           if (i > 0)
              printf("%d   %.2f\n", i, sqrt(i));
           if (i < 0)
              printf("%d   %.2fi\n", i, sqrt(abs(i)));
           }
   }
   if ((beg > end) && (end < 0)) {
      for (i = beg; i >= end; i += step) {
           if (i > 0)
              printf("%d   %.2f\n", i, sqrt(i));
           if (i < 0)
              printf("%d   %.2fi\n", i, sqrt(abs(i)));
           }
   }
}

-15   3.87i
-10   3.16i
-5   2.24i
5   2.24
10   3.16
15   3.87
20   4.47
25   5.00
30   5.48

how would i go about outputting the numbers on the right to align up

[Zhuge]

Place the width you want the number to take up after the percent sign but before the decimal point in your format string; that is a width that it will try to align to. See what happens.