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01_sizeof.pdf
This document discusses defining a macro called SIZEOF to return the size of a data type without using the sizeof operator. It explains that by taking the difference of the addresses of a variable and the variable plus one, cast to char pointers, you can get the size in bytes. An example is provided using an integer variable x, showing how taking the difference of (&x+1) and &x after casting to char pointers returns the size of an int, which is 4 bytes. Background on macros and pointers is provided. The objective is stated as understanding macro usage in preprocessing.
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01_sizeof.pdf
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- 4. Assignment 1 WAP todefine a macro SIZEOF(x),without using sizeof operator.
- 5. Assignment 1 WAP todefine a macro SIZEOF(x),without using sizeof operator. Input:
- 6. Assignment 1 WAP todefine a macro SIZEOF(x),without using sizeof operator. Input: None
- 7. Assignment 1 WAP todefine a macro SIZEOF(x),without using sizeof operator. Input: None Output:
- 8. Assignment 1 WAP todefine a macro SIZEOF(x),without using sizeof operator. Input: None Output: Print size of different data types
- 9. How will youfind the size of an data type without using sizeof operator? Assignment 1
- 10. How will youfind the size of an data type without using sizeof operator? You can make use of address of respective variable. Example: int x = 5; Assignment 1 5 x 1000 1004
- 11. How will youfind the size of an data type without using sizeof operator? You can make use of address of respective variable. Example: int x = 5; Difference of 1004 and 1000 is 4, so, use these addresses to get the size of the data type i.e., (&x+1) – (&x) Assignment 1 x 1000 1004 5
- 12. Example: intx = 5; Now, what is the output when you do (&x+1) – (&x) ? &x+1 - &x = 1000 + 1 * sizeof(int) – 1000 = 1000 + 1 * 4 – 1000 = 1004 – 1000 = 4 / sizeof(int) = 4 / 4 &x+1 - &x = 1 But, we want 4 right? So, what needs to be done to get 4? Assignment 1 x 1000 1004 5
- 13. Example: intx = 5; Now, what is the output when you do (&x+1) – (&x) ? &x+1 - &x = 1000 + 1 * sizeof(int) – 1000 = 1000 + 1 * 4 – 1000 = 1004 – 1000 = 4 / sizeof(int) <- Replace sizeof(char) = 4 / 1 &x+1 - &x = 4 How can we change it to char? Assignment 1 x 1000 1004 5
- 14. Example: intx = 5; Now, what is the output when you do (&x+1) – (&x) ? &x+1 - &x = 1000 + 1 * sizeof(int) – 1000 = 1000 + 1 * 4 – 1000 = 1004 – 1000 = 4 / sizeof(int) <- Replace sizeof(char) = 4 / 1 &x+1 - &x = 4 Typecast: (char *)(&x+1) – (char *)(&x) Assignment 1 x 1000 1004 5
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- 20. Assignment 1 Pre-requisites:- ⮚ Macros ⮚Pointers Objective:-
- 21. Assignment 1 Pre-requisites:- ⮚ Macros ⮚Pointers Objective:- ⮚ To understand the concepts of Macros in preprocessing
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