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# Problem 95 solution pg 457 this problem requires

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Unformatted text preview: resses are therefore 0x800038 and 0x800034. C. Starting with the original value of 0x800040, line 2 decremented the stack pointer by 4. Line 4 decremented it by 24, and line 5 decremented it by 4. The three pushes decremented it by 12, giving an overall change of 44. Thus, at line 11 %esp equals 0x800014. D. The stack frame has the following structure and contents: +----------+ | 0x800060 | +----------+ | 0x53 | +----------+ | 0x46 | +----------+ | | +----------+ | | +----------+ | | +----------+ | | +----------+ | | +----------+ | 0x800038 | +----------+ | 0x800034 | +----------+ | 0x300070 | +----------+ 0x80003C 0x800038 0x800034 0x800030 0x80002C 0x800028 0x800024 0x800020 0x80001C 0x800018 0x800014 <-- %ebp (x) (y) <-- %esp E. Byte addresses 0x800020 through 0x800033 are unused. 708 Problem 3.17 Solution: [Pg. 143] APPENDIX B. SOLUTIONS TO PRACTICE PROBLEMS This exercise tests your understanding of data sizes and array indexing. Observe that a pointer of any kind is four bytes long. The GCC implementation of long double uses 12 bytes to store each...
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## This note was uploaded on 09/02/2010 for the course ELECTRICAL 360 taught by Professor Schultz during the Spring '10 term at BYU.

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