This question already has answers here:
Assigning a pointer to an integer
(6 answers)
Closed 4 years ago.
I learn C programming for a while and I had to create a program, which contains a function void hello() displays word 'Hello :)' and the number, how many time the function hello() was called. The code bellow displays 'Hello' but the number of function calling stays constant. I just want to know, what's wrong and why it isn't working as it should.
#include <stdio.h>
int main(void) {
void hello(int *p_number);
int number = 1, i;
int* p_number = number;
for (i = 1; i <= 10; i++){
hello(&p_number);
printf("Number in cyclus = %d\n", number);
number++;
}
return 0;
}
void hello(int *p_number){
printf("number of calling = %d, Hello :)\n", *p_number);
}
You need
int* p_number = &number;
and
hello(p_number);
at the calling site.
i.e. set p_number to the address of number. And do turn up the warning level on your compile and read them! There is a fair amount of redundancy in maintaining a pointer in hello; presumably this is for an exercise?
#include <stdio.h>
void hello(int *p_number);
int main(void) {
int number = 1, i;
int *p_number = &number;
for (i = 1; i <= 10; i++) {
hello(p_number);
printf("Number in cyclus = %d\n", number);
number++;
}
return 0;
}
void hello(int *p_number) {
printf("number of calling = %d, Hello :)\n", *p_number);
}
Set the pointer to point at the address of number, pass pointer address to function.
Related
This question already has answers here:
Can a const variable be used to declare the size of an array in C?
(5 answers)
Closed 10 months ago.
It is saying that the variable p in the final function int main needs to be a constant but I attempted to change it to a constant and it still didn't run. Any recommendations on ways to fix it?
#define _CRT_SECURE_NO_WARNINGS // Disable warnings (and errors) when using non-secure versions of printf, scanf, strcpy, etc.
#include <stdio.h> // Needed for working with printf and scanf
// defining a function to read input
void input(double* array, int p) {
// looping p times to read p values from user
for (int i = 0; i < p; i++) {
printf("Enter a value for #%d: ", i + 1);
scanf("%lf", &array[i]);
}
}
double processing(double* array, int p) {
// defining a variable to store sum of all values
double sum = 0;
// looping over p values in the array
for (int i = 0; i < p; i++) {
sum += array[i];
}
// return average = sum / size of array
return sum / p;
}
void output(double average) {
// displaying average of all the values
// displaying only one decimal point as given
printf("The average of the values is %.1lf\n", average);
}
int main(void) {
// Constant and Variable Declarations
// defining size and initializing it to 10 as mentioned
int p = 10;
// size of p
double array[p];
input(array, p);
double average = processing(array, p);
output(average);
return 0;
} // end main() '''
Maybe you could call your function so instead of making P a variable:
input(array, 10);
sorry if it doesnt work im new to c as well
I am new to the language and was trying a simple code. I wanted to try to create a loop based on pointers. but it seems like you can not promote variate location like in assembler. or i just did it wrong? and if i really can't can i force a new variadate to be born in specific location? that was my code
#include <stdio.h>
int main(void) {
int firstnumber = 1;
int *beginning = &firstnumber;
printf("%i %i \n",firstnumber,beginning);
Test1(firstnumber,beginning);
return 0;
}
Test1 (int num, int begin)
{
int reserve = num;
if(num != 100)
{
&num +=2;
num = (reserve+1);
return Test1(num, begin);
}
else
{
int assist = begin;
while(*assist != 100)
{
printf("/n \n %i %i \n /n",num,assist);
&assist += 2;
}
}
}
I know it might look ridiculous but i'm really curious
You're thinking about this backwards. Pointers are the variables you can move around, so use them for anything you want to move around.
firstnumber is an integer variable - the compiler decides where it is stored and you can't tell the compiler to rebind the name firstnumber to a different location. You can, however, move a pointer around as much as you like. So,
void Test1(int num) {
&num +=2;
num = 42;
}
is nonsense, but
void Test2(int *num) {
num += 2;
*num = 42;
}
is fine - so long as num+2 is still a valid allocated object. For example, you could call it like
int i[5];
Test2(i); /* sets i[2] = 42 */
(if you pass in an array of fewer than 3 integers you get a runtime bug rather than a compile error, as Test2 damages your stack frame or other memory it shouldn't be touching).
No, you cannot promote/change variable location.
So this:
int num;
&num +=2;
does not make sense, and will result in an error:
error: lvalue required as left operand of assignment
&num +=2;
^~
Same for:
int assist;
&assist += 2;
Your code will compile only if you modify these statements (but now the logic is changed, you should work on that), but now it will result in an infinite loop, since your function never returns in the else case:
#include <stdio.h>
void Test1 (int num, int* begin);
int main(void) {
int firstnumber = 1;
int *beginning = &firstnumber;
printf("%i %i \n",firstnumber, *beginning);
Test1(firstnumber, beginning);
return 0;
}
void Test1 (int num, int* begin)
{
int reserve = num;
if(num != 100)
{
num +=2;
num = reserve + 1;
return Test1(num, begin);
}
else
{
int assist = *begin;
while(assist != 100)
{
printf("/n \n %i %i \n /n",num,assist);
assist += 2;
}
}
}
Good luck!
OK i checked the code deeper and got several conclusions:
function variable can not inherit his ancestor adress. therefore using a function for the task is useless unless i use some static variable
I had another conclusion but i don't remember what it was. sorry and thank you for your time
This question already has answers here:
returning a local variable from function in C [duplicate]
(4 answers)
Closed 5 years ago.
#include <stdio.h>
int* createReverseArray(int *array, int size)
{
int i;
int newArray[size];
for(i=size-1; i>=0; i--)
{
newArray[i] = array[size-i-1];
}
return newArray;
}
int main(void) {
int myArray[] = {1,2,3,5,1};
int myArray2[] = {1,2,3,4,2,1};
int i;
int size = sizeof(myArray)/sizeof(int);
printf("Size: %d\n", size);
int* newArray = createReverseArray(myArray, size);
for(i=0; i<size; i++)
{
printf("%d, ", newArray[i]);
}
return 0;
}
I printed the array within the createReverseArray function and got the correct output, but when I return the pointer to the array and then try to print the results, I think it's printing pointers to each array spot? I'm not quite sure.
This returns:
Size: 5
12341002, 1, -10772231820, -1077231812, 1074400845,
newarray is an automatic local variable. It will no longer exists once function return. Returning pointer to an automatic local variable invoke undefined behaviour and in this case nothing good can be expected.
You can allocate memory diynamically and then return pointer to it
int *newArray = malloc(sizeof(int)*size);
I want to pass two arguments into void Dividing from void Assign_numbers and void Maximum. I have only learnt to pass one argument at a time. Can you please tell me what I have to do print out the following variables inside void Dividing. If it's possible, I don't want the format of my code to change drastically. Can you also show me an example, since I am a visual learner. Thanks
#include <stdlib.h>
#include <stdio.h>
#define Max 6
struct Numbers
{
double a,b,c,d,e,f;
};
void Maximum(double *ptr);
void Dividing(double Maximum, double *ptr);
void Assign_numbers()
{
struct Numbers number;
number.a=45.78;
number.b=81.45;
number.c=56.69;
number.d=34.58;
number.e=23.57;
number.f=78.35;
Maximum((double*) &number);
Dividing((double*) &number);
}
void Maximum(double *ptr)
{
int i=0;
double Maximum = ptr[0];
for(i;i<Max;i++)
{
if(ptr[i]> Maximum)
{
Maximum = ptr[i];
}
}
Dividing(Maximum);
}
void Dividing(double Maximum, double *ptr)
{
printf("%.2f", Maximum);
printf("%.2f",ptr[3]);
}
int main()
{
Assign_numbers();
return 0;
}
Use array instead of struct - shwon here with reference example
Like Joachim Pileborg said. Don't use a struct as an array. In your case use a multidimensional array.
double[10][6] numbers;
You can easily iterate through such an array like so:
#include <stdio.h>
int main () {
/* an array with 2 rows and 6 columns*/
double numbers[2][6] = {
{45.78, 81.45, 56.69, 34.58, 23.57, 78.35},
{1,2,3,4,5, 6}
};
int i, j;
/* output each array element's value */
for ( i = 0; i < 2; i++ ) {
for ( j = 0; j < 6; j++ ) {
printf("numbers[%d][%d] = %f\n", i,j, numbers[i][j] );
}
}
/* Output by reference */
for(i = 0; i < 2; i++){
for(j=0; j < 6; j++ ){
printf("numbers[%d][%d] = %f\n", i, j,*(*(numbers + i) + j));
}
}
return 0;
}
Why the current code fails
Now onto explaining how your code (does not) work and a little about how pointers work. First off:
Dividing(double Maximum, double* ptr);
Does not work in the way you think it does. "double Maximum" is a new double variable that works within the scope of Dividing and is not a variable retrieved from the function:
void Maximum(double *ptr);
If you already knew this, then you should know or at least have expected how poor the naming of your variables are(keep it lowerCamelCase).
Now lets get onto what you're trying to do. IMHO your code is completely broken unless I am noticeing something. In Assign_numbers() you want to call Dividing() using a pointer reference. In Maximum() you want to call Dividing() again, but this time sending only a value. It doesn't make it better that you have 2 separate different calls that each have one parameter. But the function has to have two parameters. Now in order to iterate through the variables in a struct - again this is not recommended and the bottom code only serves as an example.
struct Numbers
{
double a,b,c,d,e,f;
};
struct Numbers Assign_numbers()
{
struct Numbers number;
number.a=45.78;
number.b=81.45;
number.c=56.69;
number.d=34.58;
number.e=23.57;
number.f=78.35;
return number;
}
int main()
{
struct Numbers number;
number = Assign_numbers(number);
double *value = &(number.a); //take address of the first element, since a pointer always counts upwards.
int i;
/*This loops through the addresses of the struct starting from the initial address in number.a and moves upwards 5 times and hopefully ends in number.f. Seriously bad way to construct arrays*/
/*Just try replacing sizeof(number) with sizeof(double). suddenly you get all kinds of weird values because you have ended up outside of the struct*/
/*Also note that this only works when all the datatypes in the struct have a size of 8 bytes(the size of double) */
for (i = 0; i < sizeof(number) / sizeof(double); i++){
printf("[%d]: %f\n",i, value[i]);
}
return 0;
}
New working code
With all that said. This is the closest I am going to to be able to make your code work since I have no idea what you're trying to accomplish:
#include <stdlib.h>
#include <stdio.h>
#define Max 6
struct Numbers
{
double a,b,c,d,e,f;
};
void Maximum(double *ptr);
void Dividing(double *ptr);
void Assign_numbers()
{
struct Numbers number;
number.a=45.78;
number.b=81.45;
number.c=56.69;
number.d=34.58;
number.e=23.57;
number.f=78.35;
Maximum(&number.a); //You need to parse the very first address of the struct. IN this case 'a'
Dividing(&number.a);
}
void Maximum(double *ptr)
{
int i=0;
double maximum = ptr[0];
for(i;i<Max;i++)
{
if(ptr[i]> maximum)
{
maximum = ptr[i];
}
}
printf("maximum: %f", maximum);
}
/*//removed the first parameter since it was not clear what it was for and you only had function calls to this function with one parameter */
void Dividing(double *ptr)
{
printf("%.2f",ptr[3]);
}
int main()
{
Assign_numbers();
return 0;
}
I'm a java student who's currently learning about pointers and C.
I tried to make a simple palindrome tester in C using a single array and pointer arithmetic.
I got it to work without a loop (example for an array of size 10 :*(test) == *(test+9) was true.
Having trouble with my loop. School me!
#include<stdio.h>
//function declaration
//int palindrome(int *test);
int main()
{
int output;
int numArray[10] = {0,2,3,4,1,1,4,3,2,0};
int *ptr;
ptr = &numArray[0];
output = palindrome(ptr);
printf("%d", output);
}
//function determine if string is a palindrome
int palindrome(int *test) {
int i;
for (i = 0; i <= (sizeof(test) / 2); i++) {
if (*(test + i) == *(test + (sizeof(test) - i)))
return 1;
else
return 0;
}
}
The Name of the array will itself acts as a pointer to an first element of the array, if you loose the pointer then there is no means for you to access the element of the array and hence you can send just the name of the array as a parameter to the function.
In the palindrome function:
you have used sizeof(test)/2. what happens is the address gets divided which is meaningless and hence you should not use that to calculate the mid element.
sizeof the pointer will be the same irrespective of the type of address that gets stored.
Why do you copy your pointer in another variable?
int *ptr;
ptr = &numArray[0];
Just send it to you function:
palindrome(numArray);
And sizeof(test) give you the memory size of a pointer, it's not what you want. You have to give the size in parameter of your function.
int palindrome(int *test, int size){
...
}
Finally your code must look like this:
#include<stdio.h>
int palindrome(int *test, int size);
int main()
{
int output;
int numArray[10] = {0,2,3,4,1,1,4,3,2,0};
output = palindrome(numArray, 10);
printf("%d", output);
}
//function determine if string is a palindrome
int palindrome(int *test, int size) {
int i;
for (i = 0; i < size / 2; i++) {
if (*(test + i) != *(test + (size - 1) - i))
return 0;
}
return 1;
}