- Fact:
- [ ] is an
operator !!!
It's called the array selection/subscripting operator
Evidence that [ ] is an operator:
- Take a look at any table
that list the
precedence of
operators in
C/C++
Example: click here
- You will find that [ ] is among the higher priority operators
- [ ] is an
operator !!!
- As I have said so many times:
- An operator (in any programming language)
will return
something
(It can be an value (e.g.: 3 + 4) and it can be a variable (e.g.: *p)
- An operator (in any programming language)
will return
something
- Fact:
- The [ ] operator can only be applied to a reference typed value (i.e. a pointer or an address)
- Syntax of the [ ]
operator:
p [ i ] where: p is a reference typed expression i is an integer typed expressionmeaning of the [ ] operator:
p [ i ] <===> *(p + i)Note:
- You have already learned/seen about
pointer arithmentic, so
you should know what
*(p + i) mean !!!
- You can view the [ ] operator as a shorthand notation to replace 2 operaions (* and +)
- You have already learned/seen about
pointer arithmentic, so
you should know what
*(p + i) mean !!!
- Note:
- The [ ] operator can work with
a constant of
a reference type.
Example:
double x[10]; // Defines an array x == &x[0] // x is a constant reference (address) x[i] == *(x + i) // x[i] selects the i-th element in array x
- The [ ] operator can
also be applied to a
reference typed variable p
!!!
The C compiler will translate:
translates to p[ index ] ====> *(p + index)Example:
double x[10]; // Defines an array double *p; // p is a reference variable p == &x[2] // p is a variable of a reference (address) type p[i] == *(p + i) // x[i] selects the (i+2)-th element in array x !!
- The [ ] operator can work with
a constant of
a reference type.
- Common use of the
[ ] operator:
- Instead of using
(*p + i)to access the ith array element in a dynamic array, we can now use:
p[i](This looks a whole lot sexier and a lot more like an ordinary array)
- Instead of using
- Here is the previous example on
dynamic arrays, now using the
p[...] syntax:
int main(int argc, char *argv[]) { int i; double* p; // We uses this reference variable to access // dynamically created array elements p = calloc(10, sizeof(double) ); // Make double array of 10 elements for ( i = 0; i < 10; i++ ) p[i] = i; // put value i in array element i for ( i = 0; i < 10; i++ ) printf("*(p + %d) = %lf, p[i] = %lf\n", i, *(p+i), p[i] ); free(p); // Un-reserve the first array putchar('\n'); p = calloc(4, sizeof(double) ); // Make a NEW double array of 4 elements for ( i = 0; i < 4; i++ ) p[i] = i*i; // put value i*i in array element i for ( i = 0; i < 4; i++ ) printf("*(p + %d) = %lf, p[i] = %lf\n", i, *(p+i), p[i] ); free(p); // Un-reserve the second array }Output:
*(p + 0) = 0.000000, p[i] = 0.000000 (SAME value !!!) *(p + 1) = 1.000000, p[i] = 1.000000 *(p + 2) = 2.000000, p[i] = 2.000000 *(p + 3) = 3.000000, p[i] = 3.000000 *(p + 4) = 4.000000, p[i] = 4.000000 *(p + 5) = 5.000000, p[i] = 5.000000 *(p + 6) = 6.000000, p[i] = 6.000000 *(p + 7) = 7.000000, p[i] = 7.000000 *(p + 8) = 8.000000, p[i] = 8.000000 *(p + 9) = 9.000000, p[i] = 9.000000 *(p + 0) = 0.000000, p[i] = 0.000000 *(p + 1) = 1.000000, p[i] = 1.000000 *(p + 2) = 4.000000, p[i] = 4.000000 *(p + 3) = 9.000000, p[i] = 9.000000
- Example Program:
(Demo above code)
- Prog file: click here
How to run the program:
- Right click on link(s) and
save in a scratch directory
- To compile: gcc dyn-array2.c
- To run: ./a.out
- Recall:
function with an array parameter
void print( int N, double x[ ] ) { int i; for ( i = 0; i < N; i++ ) printf("%lf\n", x[i] ); } int main( int argc, char *argv[ ] ) { double a[10]; print( a ) ; }
- Fact:
- The data type of
a in the example is:
(double *) // address of the double variable &a[0]
- The data type of
a in the example is:
- We can re-write the
definition of the function as follows:
void print( int N, double *x ) { int i; for ( i = 0; i < N; i++ ) printf("%lf\n", x[i] ); // x[i] <==> *(x + i) !!! } int main( int argc, char *argv[ ] ) { double a[10]; print( a ) ; }
- Fact:
- The above syntax equivalence
does not generalize
to more than one-level of [ ]
- In other words:
p[i][j] is invalid
- Multi-dimensional arrays:
- If you want dynamic multi-dimensional arrays in C, you have to maintain the array yourself !!!
- The above syntax equivalence
does not generalize
to more than one-level of [ ]
- Implementing a 2-dimensional
(dynamic) array:
- You must
map the
elements in a
two-dimensional array onto
a one-dimensional array.
Example:
- The mapping is as follows:
mapped to a[i][j] ===> a[ i + j*columnSize ]
- You must
map the
elements in a
two-dimensional array onto
a one-dimensional array.
- Example: a
3x4 (dynamically allocated) array
#define NRows 3 #define NCols 4 int main(int argc, char *argv[]) { int i, j; double* p; // We uses this reference variable to access // dynamically created array elements p = calloc(NRows*NCols, sizeof(double) ); // 3x4 = 12 elements for ( i = 0; i < NRows; i++ ) for ( j = 0; j < NCols; j++ ) p[i*NCols+j] = i+j; // put value i+j in array element p[i][j] for ( i = 0; i < NRows; i++ ) { for ( j = 0; j < NCols; j++ ) printf("p[%d][%d] = %lf ", i, j, p[i*NCols+j] ); putchar('\n'); } free(p); // Un-reserve the first array }Result:
p[0][0] = 0.000000 p[0][1] = 1.000000 p[0][2] = 2.000000 p[0][3] = 3.000000 p[1][0] = 1.000000 p[1][1] = 2.000000 p[1][2] = 3.000000 p[1][3] = 4.000000 p[2][0] = 2.000000 p[2][1] = 3.000000 p[2][2] = 4.000000 p[2][3] = 5.000000
- Example Program:
(Demo above code)
- Prog file: click here
How to run the program:
- Right click on link(s) and
save in a scratch directory
- To compile: gcc dyn-array3.c
- To run: ./a.out