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Sep 23 2007, 06:45 PM
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#1
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Advanced Member ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 126 Joined: 21-February 06 Member No.: 18,973 |
I'm looking for some brainstorming help. The task is to create a program where the user enters two integers, and have them multiply without using the multiplication function. For example, if the user enters 5 and 10, the program is supposed to go 10+10+10+10+10, not 5*10. I have gotten as far getting input from the user. I am thinking of using a while loop for this. Does anyone have suggestions as to the best way to do this with the while statement? Thanks.
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Sep 23 2007, 07:25 PM
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#2
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Define:EVIL PROGRAMMER (ē'vəl prō'grăm'ər)- n. An organism that converts caffeine into evil software. ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: [HOSTED] Posts: 1,023 Joined: 25-September 05 From: L.A. Member No.: 12,251 |
I think a for loop would be easier:
CODE int total=0;
for(int i=0;i<firstnumber;i++) total+=secondnumber; |
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Sep 23 2007, 07:34 PM
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#3
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Advanced Member ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 126 Joined: 21-February 06 Member No.: 18,973 |
I am not familiar with or loops. Perhaps you could describe the best method to use with a while loop?
Thanks for your help. |
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Sep 23 2007, 08:25 PM
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#4
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Define:EVIL PROGRAMMER (ē'vəl prō'grăm'ər)- n. An organism that converts caffeine into evil software. ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: [HOSTED] Posts: 1,023 Joined: 25-September 05 From: L.A. Member No.: 12,251 |
Sure it just takes a little bit more code for the same results:
CODE int total=0;
int i=0; while(i<firstnumber){ total+=secondnumber; i++; } |
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Sep 23 2007, 08:38 PM
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#5
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Advanced Member ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 126 Joined: 21-February 06 Member No.: 18,973 |
Sorry to keep perpetuating this - you've been very helpful - what is that programming in C, not C++?
Thanks for you help, A fellow OS X geek |
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Sep 23 2007, 08:49 PM
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#6
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Define:EVIL PROGRAMMER (ē'vəl prō'grăm'ər)- n. An organism that converts caffeine into evil software. ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: [HOSTED] Posts: 1,023 Joined: 25-September 05 From: L.A. Member No.: 12,251 |
It actually should work in both c and c++, and even java if you declare those variables inside a method. Most programming languages have extremely similar syntax.
This post has been edited by alex7h3pr0gr4m3r: Sep 23 2007, 08:52 PM |
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Sep 23 2007, 10:27 PM
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#7
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Advanced Member ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 126 Joined: 21-February 06 Member No.: 18,973 |
Sorry Alex - I am very new at this. Here is what I have:
CODE main () { int num1, num2, num3; printf("Please an integer:\n"); num1=GetInteger(); printf("Please a second integer:\n"); num2=GetInteger(); while (num2>1) { } getchar (); return 0; } How do I apply the code you gave me to this base? |
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Sep 24 2007, 02:45 AM
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#8
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Define:EVIL PROGRAMMER (ē'vəl prō'grăm'ər)- n. An organism that converts caffeine into evil software. ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: [HOSTED] Posts: 1,023 Joined: 25-September 05 From: L.A. Member No.: 12,251 |
try this... num3 is the result of num1 times num2
CODE int num1, num2, num3=0;
printf("Please an integer:\n"); num1=GetInteger(); printf("Please a second integer:\n"); num2=GetInteger(); int i=0; while(i<num1){ num3+=num2; i++; } getchar (); return 0; } |
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Sep 24 2007, 06:02 PM
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#9
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Advanced Member ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 126 Joined: 21-February 06 Member No.: 18,973 |
Thanks
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