i made a simple sorting program in which i initialized array like below.
int a[]={9,4,7,8,5,2,6,1,0,3};
but my sorting function sorts array a[0] to a[7] and treats 0 like '\0' and stops. if i put 0 at some other place it is sorting just up to 0 and ignores rest of the array. Is C is treating 0 and '\0' same here?
My Bubble Sort Program is as below.
#include<stdio.h>
#include<conio.h>
int main()
{
int a[]={9,4,7,8,5,2,6,1,0,3};
void bubble_sort(int *a);
void print(int *a);
bubble_sort(a);
print(a);
getch();
return 0;
}
void bubble_sort(int *a)
{
int i=0,j,t,n;
for(i=0;a[i]!='\0';i++)
{
n=0;
for(j=1;a[j]!='\0';j++)
{
if(a[j-1]>a[j])
{
t=a[j-1];
a[j-1]=a[j];
a[j]=t;
n++;
}
}
if(n==0)
{
break;
}
}
}
void print(int a[])
{
int i=0;
for(i=0;a[i]!='\0';i++)
{
printf("%d ",a[i]);
}
printf("\n");
}
'\0' is 0 by definition. You need to pass the length of your array to your sorting function, or else choose a different integer value as a terminator and make sure you never use that value for anything else.
The character '\0' is exactly equal to 0. You will want to pass an array length into your sorting algorithm, then use that.
'\0' is NULL character.ASCII code of NULL Character is 0.
a[i]!='\0' is equivalent to a[i]!=0
since '\0' is type casted to its ASCII value(Integer).
Thus the loop stops when ever current element is zero.
In order to fix it,pass an extra argument : array size
bubble_sort(a,sizeof(a)/sizeof(int));
print(a,sizeof(a)/sizeof(int));
Change the bubble_sort function to:
void bubble_sort(int *a,int sz)
{
int i=0,j,t,n;
for(i=0;i<sz;i++)
{
n=0;
for(j=1;j<sz;j++)
{
...................................
Change the print function to:
void print(int a[],int sz)
{
int i=0;
for(i=0;i<sz;i++)
{
.......................
Please note:
in C++ it is nearly impossible to calculate the size of array from decaying pointer(int *a or int a[]).
Therefore, always an extra argument : size of array is passed along with array.
However, whenever character array is passed to a function, simply iterating the array until NULL character is encountered works as in C/C++, a string of characters is stored in successive elements of a character array and terminated by the NULL character.
Is C is treating 0 and '\0' same here?
Yes. It's because both are zero. As others mentioned,or you pass the length of your array to function or use chose another delimiter for your array where you need to ensure that it will not contains in your array.
EDIT: #Jim Balter,thanks for clarification.
Related
I am learning recursion. As an exercise I am trying to find the maximum of an array recursively.
int recursive (int *arr, int n, int largest) {
if(n==0)
return largest;
else {
if (*(arr+n)>largest) {
largest = *(arr+n);
recursive(arr, n-1, largest);
}
}
}
int main() {
int length = n-1;
int largest = v[0];
int z = recursive(arr, length, largest);
printf("\n%d", z);
}
I followed your suggestions, using pointers instead of arrays, and probably the program looks way better. But still it is not doing it's not showing the maximum correctly. I think the logic is correct.
First thing pay attention to compiler warnings, your recursive function doesn't return value when you enter the else part.
Now the second thing is please don't use things like *(arr+n) which is hard to read instead use arr[n], also while just a preference when using arrays as function arguments use int arr[] to call the function instead of int *arr (in the first version it's clear you should pass an array).
Third thing is to name your things instead of int recursive describe what the function is doing for example int maxElemRecursive
So your recursive function should be something like
int maxElemRecursive(int arr[],int n,int largest)
{
if(n==0) return largest;
if(arr[n] > largest) // No need for else because return largest; would've returned value;
{
largest = arr[n];
}
return maxElemRecursive(arr,n-1,largest); // You have to return the value of the function.
// You still pass the array with just arr.
}
In C usually you can't declare an array whose size is unknown at compile-time, hence int v[n] is dangerous code.
Depending on your compiler and the compiler's settings this could be a compile error or it could be a bug.
For such problems you need to learn about pointers and dynamic memory allocation.
Side-note: After C99 there are stuff like Variable Length Arrays but the rules are a little advanced.
Also to pass an array to a function you give the array a pointer as an argument:
int z = recursion(v, n, v[0]);
instead of:
int z = recursion(v[n], n, v[0]);
I'm new to programming and I don't really understand this question. Can some of you give me examples of what it means. How do I write a function where a is int values and n is the length?
I'm confused...
I'm not sure what your question is, as you haven't provided much information. However, a function in C is defined like this:
return_type function_name( parameter list ) {
body of the function
}
So, for this situation, we could say:
void arrayFunction( int a[], int n){
//do whatever you need to do with the function here
}
This may help you some.
Suppose you have an array of ints, as follows:
int arr[] = {2,3,4,5,6};
You can see that there are 5 elements inside above array arr. You can count them.
But it happens that when you pass the above arr to function, that function has no idea about how many elements arr contains. See below (incorrect) code snippet:
#include <stdio.h>
void display(int arr[]){
}
int main(void) {
int arr[] = {2,3,4,5,6};
display(arr);
return 0;
}
The function named 'display()' has no idea about how many elements arr has
Therefore you you need to pass the extra argument (the extra argument called 'n') to tell that called function about the number of elements inside arr. You need to tell this separately - the length of arr.
Now this becomes - as you said in your question - arr is int values and n is the length
Below is the correct code:
#include <stdio.h>
void display(int a[], int n){
//Now display knows about lenth of elemnts in array 'a'
// Length is 5 in this case
}
int main(void) {
int arr[] = {2,3,4,5,6};
display(arr, 5);
return 0;
}
Now, the function named 'display()' knows the length of array of int. This is the way you write code where you specify your array and its length.
More formally, this is because while passing array, it decays to a pointer and so the need arises to pass its length also alongwith it.
I'm currently trying to pass an array of (values[3]) which it's first 3 values contain user input. However, I'm getting the error "expected int* but argument is type of int". I've tried to pass to method1, without the iteration of 'i', using the first three positions in the values array, but that's as far as I managed to attempt to fix it, any help would be much appreciated!
int main(void)
{
int i;
int values[3];
printf("Enter three consecutive numbers (With spaces between)");
scanf("%d %d %d",&values[0],&values[1],&values[2]);
for(i=0;i<3;i++)
method1(values[i]);
}
int method1(int values[3])
{
}
You cannot pass an array to a function as an array - it "decays" to a pointer. There are two issues with your code:
There is no forward declaration of method1 visible at the point of invocation, and
You are passing values[i], a scalar value in place of an array.
A forward declaration is necessary because otherwise the compiler would assume that method1 takes an returns an int, which is not true. Add this line before main
int method1(int values[]);
You could also move method1 above main to fix this without providing a forward declaration. Also, 3 inside square brackets is not necessary, because the array is passed like a pointer anyway.
If you want to pass the entire array, pass values. Of course, i becomes unnecessary:
int res = method1(values);
#include <stdio.h>
int main(void)
{
int i;
int values[3];
printf("Enter three consecutive numbers (With spaces between)");
scanf("%d %d %d",&values[0],&values[1],&values[2]);
method1(values, 3);
getchar();
getchar();
return 0;
}
int method1(int* values, int size)
{
int i;
for(i=0; i<size; i++){
printf("%d ", values[i]);
}
return 1;
}
In C, arrays are passed by reference, you can see method1's first argument is int* that refers first element of array, and second argument is size of this array.
I am trying to implement INSERTION SORT here in C, which I think I've done successfully. However, I am having problems in passing arrays as arguments.
I want in place sorting, which means that the original array passed into the insertion_sort function should contain the elements in the sorted array itself.
#include<stdio.h>
int * insertion_sort(int *a,int length)
{
int j;
for(j=1;j<length;j++)
{
int i,key=a[j];
for(i=j-1;j>=0;j--)
{
if(a[i]<=key)
break;
a[i+1]=a[i];
}
a[i+1]=key;
}
return *a;
}
int main(void)
{
int a[]={10,12,7,6,9,8};
insertion_sort(a,6);
int i;
for(i=0; i<6; i++)
printf("%d\n", a[i]);
return 0;
}
EDIT
Nothing gets printed in the output screen.
Help me find the bug here. Thanks !!
1.You probably meant to use i in the inner loop:
Change:
for(i=j-1;j>=0;j--)
^^ ^^
to:
for(i=j-1;i>=0;i--)
2.You don't have to return anything as the original array gets modified (which is just as well since you ignore the returned value).
3.Array index starts from 0. Change outer loop to: for(j=0;j<length;j++)
I have tried (sizeof(array)/sizeof(array[0])). Didn't work.
I wrote a simple function:
int length(int array[]){
int i=0;
while(array[i]) i++;
return i;
}
Worked one minute, didn't work the next.
Someone please help! I'm using Xcode as an IDE
The length of an array is not part of the array in C, so when passing an array as a parameter to a function you should pass its length as a parameter too. Here's an example:
#define ARRLEN(a) (sizeof(a)/sizeof (a)[0]) /* a must be an array, not a pointer */
void printarray(int* a, int alen)
{
int i;
for (i = 0; i < alen; i++)
printf("%d\n", a[i]);
}
main()
{
int a[] = { 3, 4, 5 };
printarray(a, ARRLEN(a));
return 0;
}
However, if your array is defined in such a way as to always end with a sentinel that isn't normal data, then you can traverse the elements until you encounter the sentinel. e.g.,
void printstrings(char** a)
{
int i;
for (i = 0; a[i]; i++)
printf("%s\n", a[i]);
}
main()
{
char* a[] = { "This", "should", "work.", NULL };
printstrings(a);
return 0;
}
Passing an array into a function is the same as passing a pointer to the array.
So sizeof(array)/sizeof(array[0]) does not work.
You can define a global macro:
#define A_LEN(a) (sizeof(a)/sizeof(a[0]))
Try this
main()
{
int a[10] ;
printf("%d\n", sizeof(a)/sizeof(int) ) ;
}
output : 10
See Why doesn't sizeof properly report the size of an array when the array is a parameter to a function? from the comp.lang.c FAQ to understand why the sizeof method doesn't work. You probably should read the entire section on arrays.
Regarding your length function, your "simple" function can work only if your array happens to have the semantic that it is terminated with an element that has the value of 0. It will not work for an arbitrary array. When an array has decayed into a pointer in C, there is no way to recover the size of the array, and you must pass along the array length. (Even functions in the C standard library are not immune; gets does not take an argument that specifies the length of its destination buffer, and consequently it is notoriously unsafe. It is impossible for it to determine the size of the destination buffer, and it therefore cannot prevent buffer overflows.)
There are several methods to get the length of array inside a function (can be in the same file or in different source file)
pass the array length as a separate parameter.
make array length as global extern so that function can directly access global data