--- title: "Pointers in pass by reference; 2D array" subject: "Computer Programming" chapter: 9 key: "semester_1/computer_programming:pointer-passing-by-reference-2d" source: https://www.bctengineer.com/q/semester_1/computer_programming/pointer-passing-by-reference-2d site: BCT Engineering --- # Pointers in pass by reference; 2D array Computer Programming · Chapter 9: Pointers - Asked in: 2068 Shrawan, 4+4 marks - Video: [Pointers and 2-D arrays](https://www.youtube.com/watch?v=sHcnvZA2u88) — mycodeschool - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/computer_programming/pointer-passing-by-reference-2d - Whole chapter: https://www.bctengineer.com/md/semester_1/computer_programming/chapter-9 ## Question What is the role of pointers in passing parameters to functions by reference? Give example, how 2 dimensional array can be accessed by a pointer. ## Short answer Passing by reference in C uses pointers: pass the address of a variable, and the function dereferences it to modify the original. For a 2D array, a pointer to an array of columns (e.g., int (*p)[3]) can access elements as *(*(p+i)+j), or a flat pointer with row-major offset. ## Answer #### Role of pointers in passing parameters by reference C passes arguments by value, so a function normally changes only copies. If the **address** of a variable is passed to a function whose parameter is a pointer, the function can reach the original variable with `*` and change it. This is how call by reference is done in C. It also allows a function to return more than one value, and it avoids copying large arrays or structures (an array name is already an address). ```c void swap(int *a, int *b) { int t = *a; *a = *b; *b = t; } ... swap(&x, &y); /* x and y are really exchanged */ ``` #### Accessing a 2D array by a pointer A 2D array is stored row by row. With `int m[2][3]`: - `m` points to row 0; `m + i` points to row i. - `*(m + i)` is row i (an array of 3 ints) and `*(*(m + i) + j)` is the element m[i][j]. - With a pointer to a row, `int (*p)[3] = m;`, the same expression `*(*(p + i) + j)` works. - Or use a flat pointer `int *q = &m[0][0];` where element [i][j] is `*(q + i*3 + j)`. ```c #include void swap(int *a, int *b) /* pass by reference */ { int t = *a; *a = *b; *b = t; } int main(void) { int x = 4, y = 9; int m[2][3] = {{1, 2, 3}, {4, 5, 6}}; int (*p)[3] = m; /* pointer to an array of 3 ints */ swap(&x, &y); printf("After swap: x = %d, y = %d\n", x, y); for (int i = 0; i < 2; i++) { for (int j = 0; j < 3; j++) printf("%d ", *(*(p + i) + j)); /* same as m[i][j] */ printf("\n"); } int *q = &m[0][0]; /* flat access */ printf("m[1][2] = %d = *(q+5) = %d\n", m[1][2], *(q + 5)); return 0; } ``` Output: ```text After swap: x = 9, y = 4 1 2 3 4 5 6 m[1][2] = 6 = *(q+5) = 6 ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can a function change the caller's variable when its address is passed? A. C passes arguments by reference by default B. The function reaches the original variable using * on the pointer parameter C. The compiler copies back the value D. Pointers are global Answer: B. C passes by value, but a passed address lets the function modify the original through *. (Key point: Role of pointers) 2. With int m[2][3], what is *(m + i)? A. The element m[i][0] value only B. Row i, an array of 3 ints C. The address of the whole array D. The element m[0][i] Answer: B. m + i points to row i, and dereferencing gives that row (an array of 3 ints). (Key point: Row of 2D array) 3. Which expression equals m[i][j]? A. *(m + i + j) B. *(*(m + i) + j) C. *(m + i) + j D. **(m + i + j) Answer: B. *(m + i) is row i, adding j moves to column j, and the outer * gives the element. (Key point: Element via pointer) 4. With int *q = &m[0][0]; and 3 columns, element [1][2] is: A. *(q + 3) B. *(q + 5) C. *(q + 6) D. *(q + 2) Answer: B. Row-wise storage gives offset i*3 + j = 1*3 + 2 = 5. 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