I have an issue here. I have to make a program that checks if a credit card number is valid or not using the checksum. I'm supposed to multiply every other digit starting from the second to last digit, then add the products then add the numbers that weren't multiplied to that sum as well. The result should produce 0 when divided. However, Im having an issue here with my program. When I enter large numbers, the values change up and at the end I get Floating point exception(core dumped). When I enter smaller numbers, sometimes it works, sometimes it doesn't. Please help me out.
Thank you for your help. Please explain the issue so I can avoid it later.
I think you may have a problem when iterating i up to a large cardNum because cardNum is long long which can hold huge numbers but int i is just an int relatively small.
To solve this problem try holding the cardNum as a string extracting each digit from it and parsing them into an int. You can then multiply and add them up without dealing with representing huge numbers.
cardnum needs to be a string not a number, you are looping over it expecting to get each digit. YOur for loop will give you all the number from 1 to the credit card number (a long loop)
Related
im looking for a base conversion function in c that could do conversions from bases 2 up to 36, including bases with characters A-Z.
For now i just found on the web functions that deal with base 2, ten and hex and a bit limited.
For this project, it would probably help to understand how bases work. In any case, let's walk through a process for how one might convert to, say, base twelve. This should be the simplest method to implement.
First up, we have our decimal number, since that's an easy place to start. Let's say, I dunno, 1452 is our number. We'll also need an array of characters for what each character is, since that'll be a lot easier than a straight ASCII conversion, where the number characters and letter characters are separated.
int dec=1452;
int toBase=12;
char outputs[36]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'}
Following that, we probably will only be OUTPUTTING the result in another base - it doesn't make sense to store it multiple ways, and makes your conversion process simpler by only converting from one base to any other given. We could store the result in a character array, but again, we already have the number stored - no point.
For this method I'm going to describe, we'll need a buffer variable to keep track of our number as we convert parts of it.
int buf=dec;
Next up, we'll start counting spaces back in the base we're going to, 12, and see what each space is worth. We'll continue until we pass our number, then backtrack one. We'll also need to save what space we're on for a for loop from that to the first space later.
int space=0;
while(Math.pow(toBase,space))<buf){
space++;
}//Braces added for clarity
space--;
Now, this is the main calculation loop, where we'll output the result. Again, the original number is still stored in 'dec,' so we don't need to worry about loss of data or changing it at all.
int i;
for(i=space;i>=0;i--){//We have set up the for loop to check each space as we progress
int modResult=buf%Math.pow(toBase,i);//Gets the number that goes in this space of the resulting base number
buf-=modResult*Math.pow(toBase,i);//We have that, so take it out of the number
printf("%c",outputs[modResult]);
}
Because of the way we're doing this, going from the top space to the bottom, modResult will never be higher than the highest number our base can go in. With this, your program will output to console the resulting number. Also, keep in mind that this only outputs the number - for the purposes of storage and calculation, it's much simpler to use the built-in functions that use base 10. Furthermore, be careful that your toBase variable never goes above 36.
As a further note, I numbered the digits (spaces), from right to left, starting at zero, because the far right space is 1, represented by your base to the zeroth power. Hope this helps.
I've been searching a while for this, and couldn't find an answer. I will be appreciated if someone knows how do to that!
PROBLEM: I must code a program that will store some numbers, but I don't know how much numbers there will be! What can I do?
I was wondering if I could use timing to get things done. I mean, if 5 secs has passed and there is no input of data, then start processing these numbers. It would work, but I couldn't code this. Can someone help, please?
1) first solution:
You can ask the user to enter the number of desired element at the beginning.
2) second solution:
Keep scan numbers till you get EOF from the user. and store the input number into a linked list or a dynamically allocated array (resize your array size withe the realloc)
3) third solution
keep scan numbers with a timeout. If there is no input during the timeout then the program will consider that the user have finished input numbers and then the program stop reading from stdin. The input numbers could stored into linked list or dynamic array as indicated in the second solution. Use select() with the scanf() in order to add the timeout behaviour as indicated in this answer
I have this function in xilinx for giving output to Seven segment.
int result;
XIo_Out32(XPAR_SSG_DECODER_0_BASEADDR, result);
The function gets the int result and puts the output to seven segment as a hex value. So basicly, if i give result = 11; I would see A as a result in seven segment. To see a decimal value on sseg, one approach is to change the verilog code behind this and change the whole concept of the sseg. Another approach is to write a function that changes decimal value into a hex value. I've been searching for a good code block for this but it seems that every one of them, prints the values digit by digit with a loop. I need the whole value as a block. Unfortunately i cannot use the C++ libraries so i have primitive C code. Is there any known algorithms for converting?
Apparently, you want to convert symbol codes from ASCII to the ones from the 7-segment display character set. If so, you may create a simple mapping, maybe an array of codes indexed by ASCII character id. Then, you'll be able to call your function like:
XIo_Out32(XPAR_SSG_DECODER_0_BASEADDR, 'A');
Be careful to implement the mapping table for the whole ASCII range.
EDIT
Sorry, I've got your question wrong. You'll have to manually convert hexadecimal number to an array of decimal symbols. You may do it by dividing your number by increasing powers of 10 (10^0, 10^1, 10^2, etc) and thus get an array of remainders, which is a decimal representation of your number. You may use snprintf as H2CO3 recommends, but I would recommend against it in some of the embedded applications where RAM is limited; you may even be unable to use sprintf-like functions at all.
Need some good problems which students can think of and apply their own logic to solve them using control instructions only. The topics covered until now are basic, not even arrays are done yet. But, I want students to be perfect before proceeding to higher topics.
I tried searching for some example problems, none were as I expected / they were the ones which I already knew.
Some of which I know:
Write a program to find out the value of a^b without using built in functions.
Write a program to find out Armstrong numbers between a range.
Write a program to print binary equivalent of a number in reverse order (since arrays are not yet done, just simple logic to print the remainder and divide the number further)
Count all -ve, +ve and 0 numbers entered by user until user wishes to terminate the program.
Write a program to display all divisors of a given number.
Write a program to find if the given number is prime or not.
Check if the given number is odd or even.
Need more good logically interesting problems which would help students to build their problem solving capability.
Thanks.
PS: Please forgive me if this question is vague or not to the point coz this question has scope for vast answers and I cannot accept a single answer, I guess?
Check if number is a palindrome (1234554321)
Rewrite a function using write() to print a number in the console (similar to printf("%d", ...))
A function that writes all combinations of 2 digits starting from 12 to 89, not allowing twice the same digit, nor a different order (12, 13, ..., 19, 23, 24... : skipping 21 because it's done with 12)
A function that write all combinations of n digits (n given as a parameter from 1 to 9) with the same rules (without using arrays)
Print first 33 terms of Fibonacci-Series
Write factorial of n being input from keyboard on console.
Find hours,minutes,seconds from given seconds.(305 s = 5m + 5s ....)
Calculate dot-product and cross-product of two 2D vectors.
Find the intersecting point of two lines(m=slope, (x0,y0)=a point for each line)
Calculate sin(pi/4) with using series expansion
Print the minimum of values given from keyboard on screen.
Simulate **and** , **or** and **xor** gates.
Find projection of a vector(3D) on another vector.
Find area of a polygon(2D)
Calculate the integral of x-square between x=0 and x=3 using integration by trapezoidal rule
Find roots of: (x-square) plus (two times x) plus (one) equals (zero)
This is just a question, please do not give me the answer. I need direction:
Assume stdin is an object reference to a Scanner, and count is an int that has been initialized to 0:
Read integers from stdin counting how many integers you see in the range 0-50 inclusive.
Stop when you read an integer outside the range.
count should be updated to indicate how many integers you read before you encounter an integer outside the range.
You can use a while loop to control the range, and then just keep reading in and adding to the count. It would look something like this:
while nextInt() is in range:
add 1 to count
It's really pretty simple, but takes some time to understand why. Basically, you have count and stdin. In the while loop condition, you check the next number from stding to see if it is in the proper range, and then add to count if it is.