exercise begginer in c

[akol ompen]

Hello,i know a little bit java but i dont know if java helps me on the language c.I did what i think is right as i see is simple .Anyway i dont know if it is right or i have some faults.Please check it .My english to explain isnt so good but i will try my best
Exercise : Is given this y=f(x)=1/x^2+1.The distance [a,b] and one step h,this makes division [a,b] in equals parts with some points [a,a+h, a+2h, … , b-h, b].It should be a program to  find and appear(printf) the x and y.Using the while.a=0,b=1,h=0.1
what i did but idk if is right
#include <stdio.h> #include <stdlib.h> main() { int a=0; int b=1; double h=0.1; int x; double y; x=a; while(x<8){ y=1/x*x+1; x=x+h; printf("x=%d",&x); printf("y=%e",&y); } system("Pause"); // here could i do it  system("0");.I see in c++ that the program finished like this way

[Campbell Ritchie]

The four languages C (not c), C++, Java®, and C# look similar, but they are all different from one another; the closest similarity is between Java® and C#. So, no, knowing one language is little help towards learning another.
Your code is poorly formatted and indented. The days when people liked to cram as much C code into as little space as possible are long gone. Put spaces round your operators and indentation. Remember you need to be able to read your code. Separate out your formula into another function which returns the value.
You are not calculating the formula correctly. You are using / x * x where the / and * operators have the same precedence. In the formula x² has a higher precedence than /, which you are not reflecting in your formula.
You are also using integer arithmetic rather than floating‑point arithmetic in line 13.
You will have problems with line 14 because you are trying to add incompatible types of number.
Give your main function a return type (probably void if you pass no arguments).

[akol ompen]

akol ompen wrote:Hello,i know a little bit java but i dont know if java helps me on the language c.I did what i think is right as i see is simple .Anyway i dont know if it is right or i have some faults.Please check it .My english to explain isnt so good but i will try my best
Exercise : Is given this y=f(x)=1/x^2+1.The distance [a,b] and one step h,this makes division [a,b] in equals parts with some points [a,a+h, a+2h, … , b-h, b].It should be a program to  find and appear(printf) the x and y.Using the while.a=0,b=1,h=0.1
what i did but idk if is right
#include <stdio.h> #include <stdlib.h> void main() // is this the way you mean ? not to return a value? { int a=0; int b=1; double h=0.1; float x; //i change them to float float y; x=a; while(x<8){ y=1/((x*x)+1); // is this better? x=x+h; printf("x=%d",&x); printf("y=%e",&y); } system("Pause"); // here could i do it  system("0");.I see in c++ that the program finished like this way.This is right on the comments?

i did what i understand it .Am i wrong ?

[akol ompen]

Campbell Ritchie wrote:The four languages C (not c), C++, Java®, and C# look similar, but they are all different from one another; the closest similarity is between Java® and C#. So, no, knowing one language is little help towards learning another.
Your code is poorly formatted and indented. The days when people liked to cram as much C code into as little space as possible are long gone. Put spaces round your operators and indentation. Remember you need to be able to read your code. Separate out your formula into another function which returns the value.
You are not calculating the formula correctly. You are using / x * x where the / and * operators have the same precedence. In the formula x² has a higher precedence than /, which you are not reflecting in your formula.
You are also using integer arithmetic rather than floating‑point arithmetic in line 13.
You will have problems with line 14 because you are trying to add incompatible types of number.
Give your main function a return type (probably void if you pass no arguments).

I cant find how can i edit my posts.Here on this site can i ask about sql?

[Campbell Ritchie]

akol ompen wrote:. . .
void main() // is this the way you mean ? not to return a value?

I think that will work all right.

. . .
float x; //i change them to float

Nonononononono. Don't use floats. Use doubles.

. . .
y=1/((x*x)+1); // is this better?

Only slightly. The x² bit is better, but you are still using integer arithmetic. Also that bit of code shou‍ld be refactored into a separate function. You have changed the formula from 1 ÷ x² + 1 to 1 ÷ (x² + 1), which is different from what you posted first. Which is correct?

x=x+h;
. . .

But x is an integer and h is floating‑point. Don't try to mix the two kinds of number in arithmetic.

In lines 15‑16, why are you using &x/&y rather than x/y?

[akol ompen]

akol ompen wrote:

akol ompen wrote:Hello,i know a little bit java but i dont know if java helps me on the language c.I did what i think is right as i see is simple .Anyway i dont know if it is right or i have some faults.Please check it .My english to explain isnt so good but i will try my best
Exercise : Is given this y=f(x)=1/x^2+1.The distance [a,b] and one step h,this makes division [a,b] in equals parts with some points [a,a+h, a+2h, … , b-h, b].It should be a program to  find and appear(printf) the x and y.Using the while.a=0,b=1,h=0.1
what i did but idk if is right
#include <stdio.h> #include <stdlib.h> void main() // is this the way you mean ? not to return a value? { int a=0; int b=1; double h=0.1; float x; //i change them to float float y; x=a; while(x<8){ y=1/((x^2)+1); // is this better? x=x+h; printf("x=%d",&x); printf("y=%e",&y); } system("Pause"); // here could i do it  system("0");.I see in c++ that the program finished like this way.This is right on the comments?

i did what i understand it .Am i wrong ?

this is right 1 ÷ (x² + 1).You asking me "In lines 15‑16, why are you using &x/&y rather than x/y? " .I read theory (i dont know if is right or wrong ,that what i understand ) i see it that in c in that way they doing it.In java for example it would be system.out.println("y= "+y); and for x the same system.out.println("x= "+x); .
external question : sql (i have no idea about it ) .Can i work online with a program like ibm db2?And here on this site there is a forum about sql?

[Henry Wong]

akol ompen wrote:You asking me "In lines 15‑16, why are you using &x/&y rather than x/y? " .I read theory (i dont know if is right or wrong ,that what i understand ) i see it that in c in that way they doing it.

In terms of right or wrong, it is about what you want done, as both expressions work. You call the printf() function to print out a decimal value to be passed in. And you passed in a pointer to the x/y variables. So... basically, you printed out the memory address locations of where the the x/y variables are located. Is that what you want done?

Henry

[Campbell Ritchie]

akol ompen wrote:. . . .You asking me "In lines 15‑16, why are you using &x/&y rather than x/y? " .I read theory (i dont know if is right or wrong ,that what i understand ) i see it that in c in that way they doing it.In java for example it would be system.out.println("y= "+y); and for x the same system.out.println("x= "+x); .

I think you have misunderstood something; as Henry says, &x prints the memory location. In Java® I would prefer to write
System.out.printf("x = %3.1f. y = %8.6f%n", x, y);
since they are floating‑point numbers.

. . .And here on this site there is a forum about sql?

Yes Try the forum called JDBC...

[akol ompen]

Campbell Ritchie wrote:

akol ompen wrote:. . . .You asking me "In lines 15‑16, why are you using &x/&y rather than x/y? " .I read theory (i dont know if is right or wrong ,that what i understand ) i see it that in c in that way they doing it.In java for example it would be system.out.println("y= "+y); and for x the same system.out.println("x= "+x); .

I think you have misunderstood something; as Henry says, &x prints the memory location. In Java® I would prefer to write
System.out.printf("x = %3.1f. y = %8.6f%n", x, y);
since they are floating‑point numbers.

. . .And here on this site there is a forum about sql?

Yes Try the forum called JDBC...

thanks a lot

[Campbell Ritchie]

My pleasure
Please show us the working version of your code.

[Campbell Ritchie]

Now that a week has gone by, I have written my working version:-// file formula.c #include <stdio.h> /* * Calculates the formula 1 ÷ (x² + 1) */ double formula(double x); /* * Pretty-print function for results of 1 ÷ (x² + 1) */ void prettyPrint(double x, double y); /* * Prints out the result of the formula function repeatedly for x 1.0...8.0 */ int main(int argc, char **argv) {    double x;    for (x = 1.0; x < 8.0; x += 0.1)    {        double y = formula(x);        prettyPrint(x, y);    } } double formula(double x) {    return 1.0 / (x * x + 1.0); } void prettyPrint(double x, double y) {    printf("x = %3.1f. y = %8.6f\n", x, y); }

campbell@campbellsComputer:~/CPrograms$ gcc -o Formula formula.c
campbell@campbellsComputer:~/CPrograms$ ./Formula
x = 1.0. y = 0.500000
x = 1.1. y = 0.452489
x = 1.2. y = 0.409836
x = 1.3. y = 0.371747
x = 1.4. y = 0.337838
x = 1.5. y = 0.307692
x = 1.6. y = 0.280899
x = 1.7. y = 0.257069
x = 1.8. y = 0.235849
x = 1.9. y = 0.216920
x = 2.0. y = 0.200000
x = 2.1. y = 0.184843
x = 2.2. y = 0.171233
x = 2.3. y = 0.158983
x = 2.4. y = 0.147929
x = 2.5. y = 0.137931
x = 2.6. y = 0.128866
x = 2.7. y = 0.120627
x = 2.8. y = 0.113122
x = 2.9. y = 0.106270
x = 3.0. y = 0.100000
x = 3.1. y = 0.094251
x = 3.2. y = 0.088968
x = 3.3. y = 0.084104
x = 3.4. y = 0.079618
x = 3.5. y = 0.075472
x = 3.6. y = 0.071633
x = 3.7. y = 0.068074
x = 3.8. y = 0.064767
x = 3.9. y = 0.061690
x = 4.0. y = 0.058824
x = 4.1. y = 0.056148
x = 4.2. y = 0.053648
x = 4.3. y = 0.051308
x = 4.4. y = 0.049116
x = 4.5. y = 0.047059
x = 4.6. y = 0.045126
x = 4.7. y = 0.043309
x = 4.8. y = 0.041597
x = 4.9. y = 0.039984
x = 5.0. y = 0.038462
x = 5.1. y = 0.037023
x = 5.2. y = 0.035663
x = 5.3. y = 0.034376
x = 5.4. y = 0.033156
x = 5.5. y = 0.032000
x = 5.6. y = 0.030902
x = 5.7. y = 0.029860
x = 5.8. y = 0.028868
x = 5.9. y = 0.027925
x = 6.0. y = 0.027027
x = 6.1. y = 0.026171
x = 6.2. y = 0.025355
x = 6.3. y = 0.024576
x = 6.4. y = 0.023832
x = 6.5. y = 0.023121
x = 6.6. y = 0.022442
x = 6.7. y = 0.021791
x = 6.8. y = 0.021169
x = 6.9. y = 0.020572
x = 7.0. y = 0.020000
x = 7.1. y = 0.019451
x = 7.2. y = 0.018925
x = 7.3. y = 0.018420
x = 7.4. y = 0.017934
x = 7.5. y = 0.017467
x = 7.6. y = 0.017018
x = 7.7. y = 0.016586
x = 7.8. y = 0.016171
x = 7.9. y = 0.015770
x = 8.0. y = 0.015385

I shall leave you to work out why you get 71 lines of output rather than 70.

[Edit]I discovered I can paste 1 ÷ (x² + 1) into line 33 and get it to print
“x = 7.0. 1 ÷ (x² + 1) = 0.020000” etc.