I wrote a program which returns a determinant of a matrix. I have a problem, actually; it always returns the "0" value. I notice that my determinant always stays as 0 even though I add numbers to it.
I wrote an English translation in the comments to understand my program better. I use a method in which we select one number and then crossed the element from the column and line of the selected number and then calculate the determinant of the uncrossed elements.
#include<stdio.h>
#include<stdlib.h>
float wznmacierz(float*macierz, int rozmiar)/*"macierz" means a matrix and "rozmiar" is a size of matrix */
{
if (rozmiar == 1)
return *macierz;
float *podmacierz = malloc((rozmiar-1)*(rozmiar-1)*sizeof(float)); // making a second matrix for uncrossed elements.
int wyznacznik = 0; // wyznacznik is the determinant of matrix
for(int element_S = 0; element_S <rozmiar; element_S++) //element s is a number from first line
{
for (int w = 1 ; w < rozmiar; w++ ) //line of checking element
{
for(int kolumna = 0; kolumna < rozmiar; kolumna++)//column of chcecking element
{
if(kolumna == element_S)
continue;
*podmacierz = macierz[(rozmiar*w)+(kolumna)];
podmacierz++;
}
}
wyznacznik += macierz[element_S]*( element_S % 2 ? -1: 1)* wznmacierz(podmacierz, rozmiar-1);
}
return wyznacznik;
}
void main()
{
float a[2][2]={{1,3},{9,8}};
printf("%d", wznmacierz(a,2));
}
Change void main to int main, because main returns an int.
In printf("%d", wznmacierz(a,2)); , change %d to %g, because %d is for formatting an int, but wznmacierz returns a float. %g will format a float. Also add \n after %g to complete the line being output.
In printf("%d", wznmacierz(a,2));, change a to *a because wzmacierz expects a pointer to a float, not a pointer to an array of float. This is a kludge to get your program “working” quickly; see Notes below.
You cannot use podmacierz both to hold the start address of the allocated array and to increment to places within the array. Inside the loop on element_S, put float *p = podmacierz; to make a second pointer, and change the uses of podmacierz inside that loop to p.
Before returning from the function, use free(podmacierz); to release the allocated space.
Notes
In main, a is declared as float a[2][2]. This makes it an array of 2 arrays of 2 float. In the call wznmacierz(a,2), a is automatically converted to a pointer to its first element. That produces a pointer to an array of 2 float. However, wznmacierz is declared with a parameter float*macierz, which is a pointer to a float.
One way to fix this is to pass *a. Once a is converted to a pointer to its first element, a pointer to an array of float, then applying * produces the thing that pointer points to, an array of float. Then that array of float is automatically converted to a pointer to its first element, producing a pointer to a float. You could also write wznmacierz(&a[0][0], 2).
This produces a pointer of the correct type for wznmacierz, which then access the array by calculating element locations, using macierz[(rozmiar*w)+(kolumna)]. This nominally calculates correct addresses for the array elements, since arrays are laid out in memory contiguously, but it is bad style unless necessary, and some people might consider it not to conform to the C standard in a pedantic sense.
One fix would be to define a in main as float a[2*2] = {1, 3, 9, 8};. Then the matrix is implemented as single flat array of float everywhere it is used.
Another fix would be to upgrade wznmarcierz to use two-dimensional arrays. A number of changes are needed to do this. I have not tested them, but I think they are at least:
Change wznmacierz(a,2) to wznmacierz(2, a).
Change the declaration of wznmacierz to float wznmacierz(int rozmiar, float macierz[rozmiar][rozmiar]).
Change the use of macierz inside the function from macierz[(rozmiar*w)+(kolumna)] to macierz[w][kolumna].
Change float *podmacierz = malloc((rozmiar-1)*(rozmiar-1)*sizeof(float)); to float (*podmacierz)[rozmiar-1] = malloc((rozmiar-1) * sizeof *podmacierz);.
Remove the float *p = podmaciarz; that I told you to insert above.
Inside the loop using w, insert float *p = podmacierz[w];.
Change macierz[element_S] to macierz[0][element_S].
Change wznmacierz(podmacierz, rozmiar-1) to wznmacierz(rozmiar-1, podmacierz).
Related
I'm having trouble with the following: I want to take a large number (cca. 15-digit) and turn it into individual digits and store them in an array. I will need these digits further on in the code. The problem is, if I declare the array outside the while loop, it stores the first four values, but then I get a segmentation fault. If I declare it within the loop, the code works as I want it to, but then I don't know how to move the array out of that loop, so that I could use it further. How can I solve this? This is what I've compiled:
unsigned long long card_num;
printf("Enter your credit card number:\n");
scanf("%llu", &card_num);
int count = 0;
while (card_num != 0)
{
int digits[count]; //declaring array into which digits will be stored
digits[count] = card_num%10; // get last digit, store it in array
card_num = card_num/10; //remove last digit
printf("Digit[%i] = %i\n", count, digits[count]);
printf("Number of digits: %i\n", count);
count ++;
}
In your code, for the very first iteration
int digits[count];
count is 0, which violates the constraints mentioned in spec. As per C11, chapter 6.7.5.2,
In addition to optional type qualifiers and the keyword static, the [ and ] may delimit an expression or *. If they delimit an expression (which specifies the size of an array), the expression shall have an integer type. If the expression is a constant expression, it shall have a value greater than zero. [....]
and
If the size is an expression that is not an integer constant expression: if it occurs in a declaration at function prototype scope, it is treated as if it were replaced by *; otherwise, each time it is evaluated it shall have a value greater than zero
So, this is not a valid code.
As already mentioned, what you are doing is plain wrong.
There is several ways to solve this issue. The easiest would be to allocate an array at the beginning of your program with enough space for all your usecases with something like :
#define MAX_DIGIT 50
int digits[MAX_DIGIT];
Then you just have to check you are not going over this array by checking that count < MAX_DIGIT.
Another way would be using dynamic memory allocation using an int pointer int *digits and malloc (I let you google that) once you know the size of the array you'll need. You'll have to change a bit your code to know the number of digits before starting to get the digits as you need to allocate the memory before starting to store digits.
You could use realloc to keep a code similar to what you already have, but I wouldn't advise it as it is not efficient to realloc memory for each value that you add.
Your logic is fine.
What went wrong is that you tried to increase the length of a fixed-length array while iterating which is not allowed.
You should never change the length of a fixed-length array anywhere in the program.
However if you want to change the length of an array during runtime you must use the malloc and realloc functions to do so.
Check out this example:
//declarations
int *storeNum;
storeNum = (int *)malloc(1 * sizeof(int));
//logic
while(num != 0) {
if(i > 0)
storeNum = realloc(storeNum, (i+1) * sizeof(int));
storeNum[i] = num % 10;
num = num/10;
++i;
}
Here first I declared the array size initially as one and later incremented it using realloc function.
You also have the array size stored in i which you can use later in your code in loops.
But keep in mind that the digits will be stored in your array in reverse order.
Also, you shouldn't declare an array within a loop.
Since you have "declared" the array, each time the compiler enters the loop while iterating it will consider the array-declaration as a new array. That is all.
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I have a function called fun(int num, int * array); which takes an int and int array as its arugments. I'm trying to convert int to the array.
When i run the program i don't get the array displayed properly
int fun(int num, int*array) {
int count =0;
while(num>0) {
num/= 10;
count++;
}
array[count];
// for loop works
}
When i print the array in the program i.e. everytime I run the program i get random digits.
What this line is supposed to do ?
array[count];
Since your integer array will change in your fun function, you have to allocate the memory (by using malloc, realloc, ...).
edit : plus, you already change the value hold by "num" when you count how many digit there are in "num".
Make a copy of "num" !
edit 2 : the more i look your function, the more it seem you will have problem using it.
Fisrt, you want to explode your integer into an array of int.
Okay, but integer have range, thus meaning integer have a maximum digit.
From my memories, there are 20 digit in an 64bit integer.
So you can simply use "int array[NB_DIGIT_INT_MAX];" with "#define NB_DIGIT_INT_MAX 21".
So, allocating is not necessary AND add complexity in your code (the caller will have to free after the function call).
Second, your fun function doesn't say how many case will hold your integer.
Let's say num = 12, your array will have "array[0] = 1, array[1] = 2", but how do you know where to stop ?
If num = 2345, how do you know that only the 4 first case in your array is legit ?
There are 2 way : You have an another variable that hold the actual size of the array, or you have a special value in your array that say "it's the end" (like '\0' for char array used as string).
You can use "-1".
Let's give a try, and don't hesitate to ask question if thing are unclear (english is not my motherlanguage)
Your array is not even allocated, this can not work as expected. You are even lucky not to have a segmentation fault. If you want to add an integer to an array making it grow, you need to allocate a larger array, copy the values and add the new one to the new array and delete the previous array, keeping array variable as a pointer to the new array. Moreover, you need to pass the size of the actual array as an argument of fun.
The count variable can be global, Initialize it outside all functions like this
short count;
The whole program could be modified like below
#include<stdio.h>
#include<stdlib.h>
short count;
void fun(int num, int **ptr) {
// You need a pointer to a pointer and return type can be void
int num_copy=num;
count=0; // Setting the count to zero
while(num_copy>0) {
num_copy/= 10; // Don't modify the original number
count++;
}
//Allocate enough memory to store the digits
*ptr=malloc(count*sizeof **ptr);
if(NULL==*ptr){
perror("Can't allocate memory");
exit(1);
}
// Then do the loop part with original num
for(int i=count -1; i>=0; i--) {
(*ptr)[i] = num%10;
num /= 10;
}
// By now you have an array 'array' of integers which you could print in main.
}
int main()
{
int number = 123456789;
int *array;
fun(number,&array); // &array is of type pointer to a pointer to integers
//Now print the individual digits
printf("Individual digits are : \n");
for(int i=count-1;i>=0;i--)
printf("%d\n",array[i]);
}
Looks to me like you are converting from an integer to digits. But I don't see where your code writes anything to the array.
If the array wasn't initialized before this, that would explain why it still contains random values.
I define three arrays. Once the first one is allocated, I print out its first element which is as expected. Then, a second array ('problematic') is allocated. When I reprint the first array's first element, it has magically changed to the value I allocated the array 'problematic' with. It gets even weirder. Had I chosen not to allocate the array 'problematic' but 'working' between the two print statements, everything works fine.
What's going on?
#include<stdio.h>
int problem(double variable);
int main(){
problem(1.0);
return 0;
}
int problem(double variable){
int limit1 = 10;
int limit2 = 10;
double bin_bounds[limit1];
double problematic[limit2];
double working[limit2];
// Allocate first array
for (int bin=0; bin<limit1; bin++){
bin_bounds[bin] = variable/limit1 * (bin+1);
}
printf("%f\n\n",bin_bounds[0]); // prints 0.2 as excpected
for (int el=0;el<=limit2;el++){problematic[el]=2.;}
printf("%f\n\n",bin_bounds[0]); // prints 2.0
return 0;
}
It's array out of bound, you are allocating 10 elements with index 0 to 9 but you're accessing index 10. Your loop should only be
for (int el=0;el<limit2;el++){problematic[el]=2.;}
The big_bounds is probably allocated right after the problematic in this case. So problematic[10] is at the same memory address as big_bounds[0]
Run with valgrind and check whether you are accesing invalid memory.During declaring array init it with 0.
I was tried to print values of a float array with 7 elements. I assigned only 3 values and they are belong to float, double and integer.
Code:
#include <stdio.h>
int main(){
float array [7];
float f = 3.24;
double d = 23.5;
int i = 4;
array[0] = f;
array[1] = i;
array[2] = d;
int n = sizeof(array)/sizeof(float);
printf("Number of Elements : %d \n\n\n",n);
for(int j = 0; j < n ; j++){
printf("%.2f ,",array[j]);
}
printf("\b ");
}
I got an output as follows :
Number of Elements : 7
3.24 ,4.00 ,23.50 ,-1.#R ,96627196995476105000000000000000.00 ,96629547147269436000000000000000.00 ,0.00
I want to clear is my code correct? And why last four values are different to each other. What are these values?
The last values are known as garbage values in C.
If you do not initialize a variable explicitly in C, it's value could be anything before you explicitly assign something to it. This anything could be garbage; the language standard does not specify what it should be.
You can read this blog for How C compiler decides garbage values: http://blog.aditech.info/2008/02/how-c-compiler-decides-garbage-values.html.
First.
Yes, your code is correct (add return 0 command at the end of your main() function though). However, it needs to be more specific, which leads to your second question.
Second.
C is a language that allows a programmer to do lots of things, but it also requires the programmer to do lots of manual coding.
So, when you declare an array of 7 items, C compiler marks a region of memory to accommodate those items (in this case of the type of float). But it doesn't actually check what that region of memory contains until you explicitly assign the values. In your case the last four values (which you have not assigned yourself) are just garbage left in the region of memory marked for your array.
Once again, compiler does not clear the memory for you when you declare the array, it just marks the region of memory. It's your responsibility to assign default values to the array.
Your possible solution is to manually initialize all of the elements of your array to some default value (for example, a 0), like this:
float array[7] = {0}
Note I'm using C not C++. I'm working on a program that will take a 2d-array and count the numbers of non-spaces in a subarray. Such as lines[0][i] <-- iterating over line[0][0] - line[0][n]. However, I'm having difficulty getting the function to accept the 2d-array.
Here is my code:
pScore[0]=letters(pLines, 0);
This is the actual function. pScore[] is another array, but a 1d one. pLines is a 4 by 200 2d-array.
int letters(char line[][], int row)
{
int i = 0;
int n = n;
int results = 0;
for( i = 0; i < (sizeof(line)/sizeof(line[0])); i++ )
{
if( !isspace(line[row][i]) )
results++;
}
return results;
}
When I do this it gives me "formal parameter number 1 is not complete". If I remove the second [] from it, the code runs but gives the worng number of non-space characters.
Taking from the comments above, and your function parameter list:
int letters(char line[][], int row)
You violate one primary tenant of passing a 2D array to a function. Specifically, you must always provide the number of columns contained in the array. e.g.
int letters(char line[][5], int row)
char line[][5] is an appropriate parameter. However, whenever you pass an array as a function parameter, the first level of indirection is converted to a pointer. (you will often here this referred to as "pointer-decay", though that is a bit of a misnomer). Therefore a proper declaration that makes this clear is:
int letters(char (*line)[5], int row)
line, after conversion, is a pointer-to-an-array of 5-int. While you can pass the array as line[][5], that is not nearly as informative as (*line)[5]. Let me know if you have any questions.
Numbers Instead of Characters
It is hard to tell what is going on without seeing the remainder of your code. However, I suspect that you are confusing the numerical value and the ASCII character value for the contents of your array. (e.g. character '0' = decimal 48 (0x30 (hex), '1' = 49, 'a' = 97, etc..). See ASCIItable.com
You you pass an array to a function it decays to a pointer. That means char line[][] should really by char (*line)[SIZE_OF_SECOND_ARRAY].
It also means the sizeof trick will not work, as doing sizeof on the pointer line will just return the size of the pointer and not what it points to, you need to explicitly pass the size as an argument to the function.
You need tell function the number of columns of 2d array. Here may help you.
I am not sure if the following statement works with you
for( i = 0; i < (sizeof(line)/sizeof(line[0])); i++ )
where sizeof(line) will be 4 or something like that depends on your platform because "line" is a pointer and you get the size of pointer itself.
Correct me if I am wrong.
In this case, you should pass column number as row's.