CODEWITHSUNDEEP
cpointers
daddydollars
daddydollars

Reputation: 11

Pointers not giving correct value

I am doing a project for school and am trying to learn how to use pointers. I am trying to get them to relay the value of X, Y, and Z. However, I keep getting the wrong value. X is supposed to equal 1. Y is supposed to equal 2. Z is supposed to equal 3.

{
// Add the C statement(s) necessary to accomplish the task identified in the comments below


double A[3] = {2.718, 3.14, 2.718}; // declare an array A of 3 doubles

A[1] = 3.14;    // put the value 3.14 in the middle location of A

A[0] = 2.718;
A[2] = 2.718;   // put the value 2.718 in the first and last locations of A


printf("The first value of A is %f\n",A[0]);    // print the first value in A


printf("The last value of A is %f\n",A[2]); // print the last value in A


A[2] = A[0] + A[1]; // change the last value in A so that it is the sum of the first two and then print it


printf("The last value of A is %f\n", A[2]);
int B[4] = {10, 25, 50, 100};   // declare an array B of 4 integers with initial values 10, 25, 50, 100


int sum = B[0] + B[1] + B[2] + B[3];
printf("The sum of all four elements in B is %d\n",sum); // print the sum of all four elements in B


printf("Four elements in B in reverse order %d %d %d %d\n",B[3], B[2], B[1], B[0]); // print the four elements in B in reverse order (100, 50, 25, 10)


int X = 1;
int Y = 2;
int Z = 3;
printf("%d, %d, %d\n,",X,Y,Z);  // declare three integers, X Y and Z, assign them the values 1, 2 and 3 and print them


int *P1;    // declare three pointers to integers, P1, P2 and P3


int *P2;
int *P3;
P1 = &X;
P2 = &Y;    
P3 = &Z;    // point P1 to X, P2 to Y, and P3 to Z.


printf("%d, %d, %d\n",P1,P2,P3);    // print the values at X and Y and Z using the pointers P1 to P3
*P1 = 10;
printf("%d", P1);   // using P1 and not X, change the variable X's value from 1 to 10, then print it


return 0;
}

Upvotes: 0

Views: 335

Answers (3)

Frankie_C
Frankie_C

Reputation: 4877

A pointer is a variable that holds the address of another variable.
A pointer to int int *p will hold the address of a variable of type int.
The operator & get the address of a variable and the operator * gets the value stored at an address.
So:

int x = 1;    //An integer i=1
int *p = &x;  // a pointer to int p to which we assign the address of x
//Now suppose x memory address is 100
printf("int=%d, pointer=%d\n", *p, p);  //Will print int=1, pointer=100

The value stored at address pointed by p is retrieved using the operator * and shows its value 1, but if you put the contents of pointer variable p you get 100.
But you can't use %d to print a pointer, you must use %p.
I used it in printf only to demonstrate the problem.

Upvotes: 0

Ravi Rajput
Ravi Rajput

Reputation: 539

&X set the Address of X to P1. but,if you want to access the value stored at that address then you should you *P1.

What does *P1 means? It means value stored at this address.

P1=&X; //P1 takes address value of X;

and if you want to print value at that address then

printf("Value of X %d\n",*P1);

Upvotes: 1

Vlad from Moscow
Vlad from Moscow

Reputation: 310930

To access values pointed to by the pointers write

printf("%d, %d, %d\n",*P1,*P2,*P3);
                      ^^^^^^^^^^^^

and

printf("%d", *P1);  
             ^^^^^

the same way as you already wrote

*P1 = 10;
^^^^^^^^

Upvotes: 0

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