Pointer to an array in c
It is most likely that you would not understand this chapter until you are through the chapter related to Pointers in C.
So assuming you have bit understanding on pointers in C programming language, let us start: An array name is a constant pointer to the first element of the array. Therefore, in the declaration:
double balance[50];
balance is a pointer to &balance[0], which is the address of the first element of the array balance. Thus, the following program fragment assigns p the address of the first element ofbalance:
double *p; double balance[10]; p = balance;
It is legal to use array names as constant pointers, and vice versa. Therefore, *(balance + 4) is a legitimate way of accessing the data at balance[4].
Once you store the address of first element in p, you can access array elements using *p, *(p+1), *(p+2) and so on. Below is the example to show all the concepts discussed above:
#include <stdio.h> int main () { /* an array with 5 elements */ double balance[5] = {1000.0, 2.0, 3.4, 17.0, 50.0}; double *p; int i; p = balance; /* output each array element's value */ printf( "Array values using pointer\n"); for ( i = 0; i < 5; i++ ) { printf("*(p + %d) : %f\n", i, *(p + i) ); } printf( "Array values using balance as address\n"); for ( i = 0; i < 5; i++ ) { printf("*(balance + %d) : %f\n", i, *(balance + i) ); } return 0; }
When the above code is compiled and executed, it produces the following result:
Array values using pointer *(p + 0) : 1000.000000 *(p + 1) : 2.000000 *(p + 2) : 3.400000 *(p + 3) : 17.000000 *(p + 4) : 50.000000 Array values using balance as address *(balance + 0) : 1000.000000 *(balance + 1) : 2.000000 *(balance + 2) : 3.400000 *(balance + 3) : 17.000000 *(balance + 4) : 50.000000
In the above example, p is a pointer to double, which means it can store address of a variable of double type. Once we have address in p, then *p will give us value available at the address stored in p, as we have shown in the above example.
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