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Using the [ ] operator to implement dynamic arrays

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• Example: dynamic array using the [ ] operator
  • 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

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  • 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