An `int` variable is stored in memory starting at address `0x1000`. Assuming an `int` occupies 4 bytes, the subsequent `int` variable, if stored immediately after, would start at a... An `int` variable is stored in memory starting at address `0x1000`. Assuming an `int` occupies 4 bytes, the subsequent `int` variable, if stored immediately after, would start at address __________.
Understand the Problem
The question discusses how memory addresses are allocated to integer variables. It states an integer variable is stored at memory address 0x1000, and asks what address the next integer variabled would be stored at, given that an integer occupies 4 bytes of memory. The key here is understanding that memory addresses increment sequentially, so the next int will be located after the current int ends.
Answer
0x1004
The subsequent int variable would start at address 0x1004.
Answer for screen readers
The subsequent int variable would start at address 0x1004.
More Information
Since each int occupies 4 bytes of memory, the next int will be located 4 bytes after the starting address of the previous one. Therefore, if the first int starts at 0x1000, the next one will start at 0x1000 + 4 = 0x1004.
Tips
A common mistake is failing to account for the size of the int when calculating the address of the subsequent variable. Always consider the data type's size in memory.
Sources
- Why 4 memory address are needed for an int variable -4 bytes - stackoverflow.com
- Memory address size - C++ Forum - CPlusPlus - cplusplus.com
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