Yes, this is a basic C coding homework problem. No, I am not just looking for someone to do it for me. Considering that this is my first programming class, I'm not surprised that I can't get it to work, and I'm certain there is plenty wrong with it. I just want some help pointing out the problems in my code and the things that are missing so that I can fix them on my own.
Homework Question:
Write a program to read ONLY one integer number (your input must be
one 3 digit number from 100 to 999), and to think of a number as
being ABC (where A, B, and C are the 3 digits of a number). Now,
form the number to become ABC, BCA, and CAB, then find out the
remainder of these three numbers when they are divided by 11.
Assume remainders would respectively be X, Y, and Z and add them
up as X+Y, Y+Z, and Z+X. Now if any of these summations is odd
number, increase it by 11 if the summation plus 11 is less than 20,
otherwise decrease the summation by 11 (this summation operation
must be positive number but less than 20). Finally, divide each
of the sums in half. Now, print out all the resulting digits.
My Code:
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
int main()
{
//Declare all variables
int OrigNumber;
int x, y, z;
int number;
number = x, y, z;
int sum;
//
printf("Input a three digit number");
//
int c;
c = OrigNumber %10;
//
int b;
b=((OrigNumber - c) % 100)/10;
//
int a;
a = (OrigNumber - (b + c))/100;
//
int abc, bca, cab;
abc = (a*100) + (10*b) + c;
bca = (10*b) + c + (a*100);
cab = c + (a*100) + (10*b);
//
if((number % 2) == 1)
{
if(number + 11 < 20)
number += 11;
else if((100 - 11 > 0) && (100 - 11 < 20))
number -= 11;
}
//
x = abc/11;
y = bca/11;
z = cab/11;
//
sum = (x + y),
(y + z),
(z + x);
}
To start with, you need to read the input. Start with a prompt that includes a carriage return:
printf("Input a three digit number: \n");
Since it's a three digit number, you could add the following line to read the input:
scanf("%3d", &OrigNumber);
The next bit of code works quite well until you get to your if (number % 2) which is meaningless since you didn't really define number - well, you did, but the line
number = x, y, z;
does NOT do what you think it does. If you add
printf("So far I have abc=%d, bca=%d, cab=%d\n", abc, bca, cab);
after you first read in the number and computed those three, you will see you are well on your way.
Note that
number = x, y, z;
Uses a thing called the "comma operator". All the things (a,b,c) are "evaluated" but their values are not returned. At any rate, where you have that line, you didn't yet assign a value to x,y and z.
Is that enough to get your started?
update now that you have had a few hours to mull this over, here are a few more pointers.
Your computation of abc, cab, bca makes no sense. I will show you just one of them:
cab = c*100 + a*10 + b;
Next you need to compute each of x, y and z. Again, here is one of the three:
y = bca%11;
Now you have to make the sums - I call them xy, yz, and zx. Just one of them:
zx = z + x;
Next, to deal with the instruction: "Now if any of these summations is odd number, increase it by 11 if the summation plus 11 is less than 20, otherwise decrease the summation by 11:
if(xy % 2 == 1) {
if(xy + 11 < 20) xy += 11; else xy -= 11;
}
use similar code for all three sums. Then "divide by 2":
xy /= 2;
repeat as needed.
Finally, print out the result:
printf("xy: %d, yz: %d, zx: %d\n", xy, yz, zx);
The amazing thing is that if you did this right, you get the original numbers back...
You could make the code more compact by using an array of values and looping through it - rather than repeating the code snippets I wrote above with different variables. But I suspect that is well outside the scope of what you are expected to know at this point.
Can you take it from here?
#include <stdio.h>
int main()
{
//Declare all variables
int OrigNumber;
int a, b, c;
int abc, bca, cab;
int x, y, z;
int xplusy , yplusz, xplusz;
printf(" A program to read ONLY one integer number.\n Input must be one 3 digit number from 100 to 999 : ");
scanf("%d", &OrigNumber); // Get input from console
if(OrigNumber > 999 || OrigNumber < 100) {
printf("Invalid number. Quiting program. This is error handling. Important while learning programming.");
return 0;
}
c = OrigNumber %10; // digit at unit's place
b=((OrigNumber) % 100)/10; //digit at the ten's place
a = (OrigNumber)/100; //digit at the 100's place. Note: 734/100 = 7. NOT 7.34.
printf("\n Three numbers say A,B, C : %d, %d , %d ", a, b, c);
abc = a*100 + 10*b + c;
bca = 100*b + 10*c + a;
cab = c*100 + a*10 + b;
printf("\n Three numbers say ABC, BCA, CAB : %d, %d , %d ", abc, bca, cab);
x = abc % 11; // Reminder when divided by 11.
y = bca % 11;
z = cab % 11;
printf("\n Three numbers say X, Y, Z : %d, %d , %d ", x, y, z);
xplusy = x + y; // Adding reminders two at a time.
yplusz = y + z;
xplusz = x + z;
printf("\n Three numbers X+Y, Y+Z, X+Z : %d, %d , %d ", xplusy, yplusz, xplusz);
if((xplusy % 2) == 1) {
if(xplusy + 11 < 20)
xplusy += 11;
else
xplusy -= 11;
}
if((yplusz % 2) == 1) {
if(yplusz + 11 < 20)
yplusz += 11;
else
yplusz -= 11;
}
if((xplusz % 2) == 1) {
if(xplusz + 11 < 20)
xplusz += 11;
else
xplusz -= 11;
}
xplusy /= 2; // Finally, divide each of the sum in half.
yplusz /= 2;
xplusz /= 2;
printf("\n Now print out all the resulting digits : %d, %d , %d \n", xplusy, yplusz, xplusz);
return 0;
}
int abc, bca, cab;
abc = (a*100) + (10*b) + c;
bca = (10*b) + c + (a*100);
cab = c + (a*100) + (10*b);
I suggest printing out the numbers at this point in the code.
printf( "%d %d %d", abc, bca, cab );
I think you'll see one of the problems you need to solve.
#include <iostream>
#include <cstdio>
using namespace std;
int main() {
int n, a, b, c, abc, bca, cab, x, y, z, p, q, r;
scanf("%d", &n);
c=n%10;
b=(n/10)%10;
a=n/100;
abc=a*100+b*10+c;
bca=b*100+c*10+a;
cab=c*100+a*10+b;
x=abc%11;
y=bca%11;
z=cab%11;
p=x+y;
q=y+z;
r=z+x;
return 0;
}
Now if any of these summations is odd number, increase it by 11 if the
summation plus 11 is less than 20, otherwise decrease the summation by
11 (this summation operation must be positive number but less than
20). Finally, divide each of the sums in half. Now, print out all the
resulting digits.
i didnt get the final part, can you explain it more clearly?
Related
#include <stdio.h>
int main()
{
signed int x;
int x1 = 0, x2 = 10, final, loop = 1, y = 10, c;
printf("Enter the value of X.\n");
scanf("%d", &x);
printf("Value Scanned:%d\n", x);
again:
if (loop <= 32)
{
if (x >= x1 && x < x2)
{
final = x - x1;
printf("%d", final);
y = y * 10;
x1 = 0;
x2 = 0;
++loop;
goto again;
}
else
{
c = x2 - x1;
if (x1 == x2)
{
x2 += y;
goto again;
}
else if (c == y)
{
x1 += y;
x2 += y;
goto again;
}
else
{
printf("Error in Process");
goto ending;
}
}
}
else
{
printf("0+error, extra long input");
}
ending:
return 0;
}
Flowchart:
I am a beginner in C-language and only know how to use If-else, Switch, Goto statements, with basic knowledge of how to integrate basic level loops. So please tell me what/where I am wrong instead of telling me how to use arrays because I don't know them, etc. This is my most complex code until now.
Now for Explanation of Coding,
I wrote X1 as the lower value and X2 as the upper value while first keeping a difference = Y(initially 10) between them.
Continuously increasing the value of X1 and X2 by Y(10) together simultaneously, I will arrive in between an intersection where my x(input) lies.
Eg-
x=568
then X1 and X2 will keep on increasing until they reach X1 = 560 and X2 = 570, then they will do Final = X(568) - X1(560) and print it.
since it can only happen for 32-digits long, so I wrote loop = 0 and only processing my main statement till loop is smaller than or equal to 32, otherwise printing "0+error".
then I put Y = Y * 10 every time the value was within my specified range.
It should give me the values like Last digit, then last 2 digits, then last 3 digits,etc. but after scanning the value, it isn't exciting at all.
Evaluating what you are attempting to do, I reworked your code to make it a bit more structured without utilizing arrays at that seemed to be something you wanted to avoid at this time. However, these days, the use of the goto statement is usually avoided since functionality such as for loops, do/while loops, and while loops offer better clarity to coding. With that in mind, following is a snippet of code that provides the functionality it looks like you want.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int x, x1, x2, y = 10, counter = 0, last_digit;
printf("Please enter a number: ");
scanf("%d", &x);
if (x < 0) /* Just in case a negative integer is entered */
{
x = x * -1;
}
while (1) /* Use a while loop with associated break statements to avoid goto and label statements */
{
x1 = 0;
x2 = 10;
counter += 1;
while (1)
{
if (x >= x1 && x <= x2)
{
last_digit = x - x1;
if (counter == 1)
{
printf("The last digit is: %d\n", last_digit);
}
else
{
printf("The next digit is: %d\n", last_digit);
}
break;
}
x1 += y;
x2 += y;
}
x = x / 10; /* Perform integer division by ten to get to the next digit in the entered number */
if (x == 0) /* Once all digits have been processed the outer while loop can be exited */
{
break;
}
}
return 0;
}
Following are some key points.
As noted, the loop process using goto statements is replaced by two while loops; one while loop nested inside another while loop.
Utilizing integer division by ten, each digit can be ascertained and printed.
Utilizing the nested while loops with break statements allows for a more compact program.
Utilizing this code snippet, following is a sample test from the terminal.
#Dev:~/C_Programs/Console/LastDigit/bin/Release$ ./LastDigit
Please enter a number: 479824385
The last digit is: 5
The next digit is: 8
The next digit is: 3
The next digit is: 4
The next digit is: 2
The next digit is: 8
The next digit is: 9
The next digit is: 7
The next digit is: 4
Back in the day, the goto statement had its place in coding, but today it is pretty much an artifact.
Give that a try and see if it meets the spirit of your project.
I'm trying to code a program that can tell apart real and fake credit card numbers using Luhn's algorithm in C, which is
Multiply every other digit by 2, starting with the number’s
second-to-last digit, and then add those products’ digits together.
Add the sum to the sum of the digits that weren’t multiplied by 2.
If the total’s last digit is 0 (or, put more formally, if the total
modulo 10 is congruent to 0), the number is valid!
Then I coded something like this (I already declared all the functions at the top and included all the necessary libraries)
//Luhn's Algorithm
int luhn(long z)
{
int c;
return c = (sumall(z)-sumodd(z)) * 2 + sumaodd(z);
}
//sum of digits in odd position starting from the end
int sumodd(long x)
{
int a;
while(x)
{
a = a + x % 10;
x /= 100;
}
return a;
}
//sum of all digits
int sumall(long y)
{
int b;
while(y)
{
b = b + y % 10;
y /= 10;
}
return b;
}
But somehow it always gives out the wrong answer even though there's no error or bug detected. I came to notice that it works fine when my variable z stands alone, but when it's used multiple times in the same line of code with different functions, their values get messed up (in function luhn). I'm writing this to ask for any fix I can make to make my code run correctly as I intended.
I'd appreciate any help as I'm very new to this, and I'm not a native English speaker so I may have messed up some technical terms, but I hope you'd be able to understand my concerns.
sumall is wrong.
It should be sumeven from:
Add the sum to the sum of the digits that weren’t multiplied by 2.
Your sumall is summing all digits instead of the non-odd (i.e. even) digits.
You should do the * 2 inside sumodd as it should not be applied to the other [even] sum. And, it should be applied to the individual digits [vs the total sum].
Let's start with a proper definition from https://en.wikipedia.org/wiki/Luhn_algorithm
The check digit is computed as follows:
If the number already contains the check digit, drop that digit to form the "payload." The check digit is most often the last digit.
With the payload, start from the rightmost digit. Moving left, double the value of every second digit (including the rightmost digit).
Sum the digits of the resulting value in each position (using the original value where a digit did not get doubled in the previous step).
The check digit is calculated by 10 − ( s mod 10 )
Note that if we have a credit card of 9x where x is the check digit, then the payload is 9.
The correct [odd] sum for that digit is: 9 * 2 --> 18 --> 1 + 8 --> 9
But, sumodd(9x) * 2 --> 9 * 2 --> 18
Here's what I came up with:
// digsum -- calculate sum of digits
static inline int
digsum(int digcur)
{
int sum = 0;
for (; digcur != 0; digcur /= 10)
sum += digcur % 10;
return sum;
}
// luhn -- luhn's algorithm using digits array
int
luhn(long z)
{
char digits[16] = { 0 };
// get check digit and remove from "payload"
int check_expected = z % 10;
z /= 10;
// split into digits (we use little-endian)
int digcnt = 0;
for (digcnt = 0; z != 0; ++digcnt, z /= 10)
digits[digcnt] = z % 10;
int sum = 0;
for (int digidx = 0; digidx < digcnt; ++digidx) {
int digcur = digits[digidx];
if ((digidx & 1) == 0)
sum += digsum(digcur * 2);
else
sum += digcur;
}
int check_actual = 10 - (sum % 10);
return (check_actual == check_expected);
}
// luhn -- luhn's algorithm using long directly
int
luhn2(long z)
{
// get check digit and remove from "payload"
int check_expected = z % 10;
z /= 10;
int sum = 0;
for (int digidx = 0; z != 0; ++digidx, z /= 10) {
int digcur = z % 10;
if ((digidx & 1) == 0)
sum += digsum(digcur * 2);
else
sum += digcur;
}
int check_actual = 10 - (sum % 10);
return (check_actual == check_expected);
}
You've invoked undefined behavior by not initializing a few local variables in your functions, for instance you can remove your undefined behaviour in sumodd() by initializing a to zero like so:
//sum of digits in odd position starting from the end
int sumodd(long x)
{
int a = 0; //Initialize
while(x)
{
a += x % 10; //You can "a += b" instead of "a = a + b"
x /= 100;
}
return a;
}
It's also important to note that long is only required to be a minimum of 4-bytes wide, so it is not guaranteed to be wide enough to represent a decimal-16-digit-integer. Using long long solves this problem.
Alternatively you may find this problem much easier to solve by treating your credit card number as a char[] instead of an integer type altogether, for instance if we assume a 16-digit credit card number:
int luhn(long long z){
char number[16]; //Convert CC number to array of digits and store them here
for(int c = 0; c < 16; ++c){
number[c] = z % 10; //Last digit is at number[0], first digit is at number[15]
z /= 10;
}
int sum = 0;
for(int c = 0; c < 16; c += 2){
sum += number[c] + number[c + 1] * 2; //Sum the even digits and the doubled odd digits
}
return sum;
}
...and you could skip the long long to char[] translation part altogether if you treat the credit card number as an array of digits in the whole program
This expression:
(sumall(z)-sumodd(z)) * 2 + sumall(z);
Should be:
((sumall(z)-sumodd(z)) * 2 + sumodd(z))%10;
Based on your own definition.
But how about:
(sumall(z) * 2 - sumodd(z))%10
If you're trying to be smart and base off sumall(). You don't need to call anything twice.
Also you don't initialise your local variables. You must assign variables values before using them in C.
Also you don't need the local variable c in the luhn() function. It's harmless but unnecessary.
As others mention in a real-world application we can't recommend enough that such 'codes' are held in a character array. The amount of grief caused by people using integer types to represent digit sequence 'codes' and identifiers is vast. Unless a variable represents a numerical quantity of something, don't represent it as an arithmetic type. More issue has been caused in my career by that error than people trying to use double to represent monetary amounts.
#include <stdio.h>
//sum of digits in odd position starting from the end
int sumodd(long x)
{
int a=0;
while(x)
{
a = a + x % 10;
x /= 100;
}
return a;
}
//sum of all digits
int sumall(long y)
{
int b=0;
while(y)
{
b = b + y % 10;
y /= 10;
}
return b;
}
//Luhn's Algorithm
int luhn(long z)
{
return (sumall(z)*2-sumodd(z))%10;
}
int check_luhn(long y,int expect){
int result=luhn(y);
if(result==expect){
return 0;
}
return 1;
}
int check_sumodd(long y,int expect){
int result=sumodd(y);
if(result==expect){
return 0;
}
return 1;
}
int check_sumall(long y,int expect){
int result=sumall(y);
if(result==expect){
return 0;
}
return 1;
}
int main(void) {
int errors=0;
errors+=check_sumall(1,1);
errors+=check_sumall(12,3);
errors+=check_sumall(123456789L,45);
errors+=check_sumall(4273391,4+2+7+3+3+9+1);
errors+=check_sumodd(1,1);
errors+=check_sumodd(91,1);
errors+=check_sumodd(791,8);
errors+=check_sumodd(1213191,1+1+1+1);
errors+=check_sumodd(4273391,15);
errors+=check_luhn(1234567890,((9+7+5+3+1)*2+(0+8+6+4+2))%10);
errors+=check_luhn(9264567897,((9+7+5+6+9)*2+(7+8+6+4+2))%10);
if(errors!=0){
printf("*ERRORS*\n");
}else{
printf("Success\n");
}
return 0;
}
#include <stdio.h>
#include <stdlib.h>
int main()
{
int x, y, z, result, max;
printf("\nInput the first integer: ");
scanf("%d", &x);
printf("\nInput the second integer: ");
scanf("%d", &y);
printf("\nInput the third integer: ");
scanf("%d", &z);
result=(x+y+abs(x-y))/2;
max=(result+z+abs(result-z))/2;
printf("\nMaximum value of three integers: %d", max);
printf("\n");
return 0;
}
unable to understand the formula:
result=(x+y+abs(x-y))/2;
max=(result+z+abs(result-z))/2;
Looking at this expression:
(x+y+abs(x-y))/2
If x => y, then abs(x-y) is the same as x-y. That gives us: (x+y+(x-y))/2 == (x+x+y-y)/2 == 2x/2 == x.
If x < y, then abs(x-y) is the same as y-x. That gives us: (x+y+(y-x))/2 == (x-x+y+y)/2 == 2y/2 == y.
So the above expression evaluates to the larger of x and y without using any conditionals. The following expression (result+z+abs(result-z))/2 does the same thing with z and the max of x and y.
Note however that this method has the potential to cause overflow. The cleanest way to do this is to explicitly compare:
if (x >= y && x >= z) {
max = x;
} else if (y >= x && y >= z) {
max = y;
} else {
max = z;
}
then how to solveit – Ujjwal Bhardwaj 5 mins ago
int max(int a, int b, int c)
{
return a > b ? (a > c ? a : c) : (b > c ? b : c);
}
One way to visualize it is:
Imagine you have 2 trees. One is 16 meter tall and the other is 20 meter tall.
You look at their average, which is the “midpoint” and is 18 meter tall. Now, what’s their difference? 4 meter.
You take 18 and add half of that difference, is 20 (the max). Likewise, you can take the average and minus half of the difference and it is the min.
So,
average plus half the difference
= (x + y) / 2 + abs(x - y) / 2
= (x + y + abs(x - y)) / 2
The following line of code returns the value of whichever is greater, x or y.
result = (x + y + abs(x - y)) / 2;
By adding the absolute value of the difference between x and yto the sum of x and y you essentially get 2 times the larger number. For example, if x=5 and y=20 then abs(x - y) = 15. So 5 + 20 + 15 = 40 which is 2 times the larger number. Divide that by 2 and you have determined larger value. Then by repeating the formula with the result above and z you have calculated the largest of the three.
I am learning c. So, I was practicing in a online judge. I have got a logic behind the problem and submitted ans got wrong answer. What is the problem?
Problem:
Area
100 Points · Limits 1s, 512 MB
In this problem, you will be given a square which has a length of n. Co-ordinates of the square are (0,0), (n,0),(n,n),(0,n) . You need to draw 4 straight lines:
Line from (0,1) to (n,n-1)
Line from (1,0) to (n-1,n)
Line from (0,n-1) to (n,1)
Line from (1,n) to (n-1,0)
These four lines will intersect in a point (x,y) like the figure shown below.
Calculate the total area of A+B+C+D (except the four corner unit square).
Input
Input will start with an integer T. Then there will be T cases. Each case will contain one integer N. 1 <= T <= 100000
3 <= n <= 1018
Output
For each test case, print “Case x: y” without quotation marks where x is the case number and y is the required answer.
It is guaranteed that y is always an integer.
Sample
Input Output
1
6
Case 1: 8
My code:
#include <stdio.h>
#include <stdlib.h>
#include<math.h>
int main()
{
int test, i;
scanf("%d", &test);
for(i=0; i<test; i++)
{
double n, area, a,x, b1, b, s, tri, area1, area_t;
scanf("%lf", &n);
area= n*n;
a=n-2;
x=n/2;
b1= (x-1)*(x-1) + x*x;
b= sqrt(b1);
s= (a+b+b)/2;
area1= s*(s-a)*(s-b)*(s-b);
area_t = (4* sqrt(area1));
printf("Case %d: %.0lf\n",i+1, (area-(area_t + 4)));
}
return 0;
}
Please help me to improve the code. Thank you.
I believe you have a compatability problem and your online judge is on a C89 implementation where "%lf" does not exist, making your program output
Case 1: %.0lf
Case 2: %.0lf
...
Try using the C89 specifier
printf("Case %d: %.0f\n", i + 1, area - (area_t + 4));
/* ^^^^ C89, not %.0lf */
Note: double x; scanf("&lf", &x) has been valid since C89.
Let Area be the requested area, which can be calculated as:
Area = OuterSquareArea - 4 * IsoscelesTringleArea - 4 * SmallSqareArea
where:
OuterSqareArea = n * n
IsoscelesTriangleArea = base * height / 2
= (n - 2) * (n / 2) / 2
= (n - 2) * n / 4
SmallSquareArea = 1 * 1
= 1
The computation of Area can be summed up to:
Area = (n * n) - 4 * ((n - 2) * n / 4) - 4 * (1)
= n * n - (n - 2) * n - 4
= (n - (n - 2)) * n - 4
= 2 * n - 4
The trace guarantees that t, n and Area are integers. The code we need is then:
#include <stdio.h>
int main() {
int i, t, n;
scanf("%d", &t); /* read t */
for (i = 1; i <= t; i++) { /* for any i in [1,t] */
scanf("%d", &n); /* read n */
printf("Case %d: %d\n", i, 2 * n - 4); /* solve */
}
}
The code could simply be extented to check t and n to be in the given ranges.
I've written some code to Integrate the function 5x^4 + 4x^3 + 3x^2 + 2x + 1.
#include<stdio.h>
#include<math.h>
float func(float x){
float a;
a = 5*pow(x,4) + 4*pow(x,3) + 3*pow(x,2) + 2*x +1;
return a;
}
int main(){
float numberOfXValues;
float a = 0; //lower limit
float b = 1; //upper limit
float numberOfStrips;
float stripSize;
float finalAnswer;
float sumFirstAndLast; //summation of first and last x value
while(1){
printf("Input number of X values:");
scanf("%f", &numberOfXValues);
numberOfStrips = numberOfXValues - 1;
stripSize = (b - a)/(numberOfStrips);
sumFirstAndLast = 0.5*func(a) + 0.5*func(b);
for (float z = stripSize; z < b; z += stripSize ){
sumFirstAndLast += func(z);
}
finalAnswer = sumFirstAndLast * stripSize;
printf("%f\n", finalAnswer);
}
return 0;
}
And it works for the majority of values, but the output for 13 and 20 is giving the wrong answer. I've looked through it a few times but can't see what could be causing this.
Input number of X values:10
5.039070
Input number of X values:11
5.031651
Input number of X values:12
5.026160
Input number of X values:13
6.271982
Input number of X values:14
5.018732
Input number of X values:15
5.016153
Input number of X values:16
5.014071
Input number of X values:17
5.012367
Input number of X values:18
5.010955
Input number of X values:19
5.009773
Input number of X values:20
5.798243
Input number of X values:21
5.007917
The problem is that in the exit condition of the inner loop, you compare two floating point numbers. After numberOfStrips iterations, it's not guaranteed that z == b because of floating point accuracy. It's entirely possible that z < b or z > b. In the z < b case, the loop it executed another time.
You should make numberOfStrips an integer and rewrite the loop like this:
float z = stripSize;
for (int i = 1; i < numberOfStrips; i++) {
sumFirstAndLast += func(z);
z += stripSize;
}
At a first glance this looks to precision and rounding issues of float.
Therefore change float to double.
For example on my machine I get 6.271982 for an input of 10 when using float while I get 5.021983 as result when using double.
Moreover you should use an epsilon in floating point comparisons.
Placing the line:
printf("%f, %d\n", z, z < b);
Inside the for loop will print the z values each iteration along with the loop condition result. The last output of this will be:
1.000000, 1
Which is pointing that we have already reached the right limit, but still performing the loop body. Literally it is thinking that 1.000 < 1.000. It can happen when working with float numbers, so in order to prevent it, consider looping on some integer number (numberOfXValues for example).