Java Programming
9th Edition
ISBN: 9780357616635
Author: Joyce Farrell
Publisher: CENGAGE L
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Expert Solution & Answer
Chapter 4, Problem 1RQ
Program Description Answer
The “block” means the code between a pair of curly braces in a method.
Hence, the correct option is “C”.
Expert Solution & Answer
Explanation of Solution
Blocks:
- The code between a pair of curly braces within any method or class is called as blocks.
- It contains two blocks. One is outer block and another is inner block.
- The outer block starts at the first opening brace and ends at the last closing brace at the end of the method.
- The inner block begins with the second opening curly brace and ends with the first closing curly brace.
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Students have asked these similar questions
Match each of the assembler routines on the left with the equivalent C function on
the right. Write the name of the label (e.g., foo) to the right of the corresponding
function. Note: shrq is the logical right shift instruction, and sarq is the arithmetic
right shift instruction.
foo1:
leaq
0(,%rdi, 8), %rax
long choice1 (long x)
{
ret
return x -
8 >8;
foo3:
}
movq
sarq
%rdi, %rax
$8, %rax
long choice4 (long x)
ret
{
return x*256;
}
foo4:
long choice5 (long x)
leaq
-8 (%rdi), %rax
{
ret
return x-8;
}
long choice6 (long x)
foo5:
{
leaq
-8 (%rdi), %rax
return x+8;
shrq
$63, %rax
}
ret
Given the variables and code in the text below, identify where in memory they will
live once the code is compiled.
1 char
big_array [1L<<24]; /* 16 MB */
2 GB *
:/
2 char huge_array [1L<<31]; /*
3
4 int global = 0;
5
6 int useless () { return 0; }
7
8 int main()
9 {
10
void *p1, p2, *p3, *p4;
int local =
0;
malloc (1L << 28); /* 256 MB *,
11
12
p1
13
p2
=
malloc (1L << 8);
/* 256 B *
14
p3
15
p4
=
malloc (1L << 32);
malloc (1L << 8);
/* 4
GB *
*/
/* 256
B */
16 }
Note: *pN is the thing at which pN points.
1. big_array
2. huge_array
3. global
4. useless
5. void* p1
6. *p1
7. void* p2
8. *p2
9. void* p3
10. *p3
11. void* p4
12. *p4
The next problem concerns the following C code:
/copy input string x to buf */
void foo (char *x) {
char buf [8];
strcpy((char *) buf, x);
}
void callfoo() {
}
foo("ZYXWVUTSRQPONMLKJIHGFEDCBA");
Here is the corresponding machine code on a Linux/x86 machine:
0000000000400530 :
400530:
48 83 ec 18
sub
$0x18,%rsp
400534:
48 89 fe
mov
%rdi, %rsi
400537:
48 89 e7
mov
%rsp,%rdi
40053a:
e8 di fe ff ff
callq
400410
40053f:
48 83 c4 18
add
$0x18,%rsp
400543:
c3
retq
400544:
0000000000400544 :
48 83 ec 08
sub
$0x8,%rsp
400548:
bf 00 06 40 00
mov
$0x400600,%edi
40054d:
e8 de ff ff ff
callq 400530
400552:
48 83 c4 08
add
$0x8,%rsp
400556:
c3
This problem tests your understanding of the program stack. Here are some notes to
help you work the problem:
⚫ strcpy(char *dst, char *src) copies the string at address src (including
the terminating '\0' character) to address dst. It does not check the size of
the destination buffer.
• You will need to know the hex values of the following characters:
Chapter 4 Solutions
Java Programming
Ch. 4 - Prob. 1RQCh. 4 - Prob. 2RQCh. 4 - Prob. 3RQCh. 4 - Prob. 4RQCh. 4 - Prob. 5RQCh. 4 - Prob. 6RQCh. 4 - Prob. 7RQCh. 4 - Prob. 8RQCh. 4 - Prob. 9RQCh. 4 - Prob. 10RQ
Ch. 4 - Prob. 11RQCh. 4 - Prob. 12RQCh. 4 - Prob. 13RQCh. 4 - Prob. 14RQCh. 4 - Prob. 15RQCh. 4 - Prob. 16RQCh. 4 - Prob. 17RQCh. 4 - Prob. 18RQCh. 4 - Prob. 19RQCh. 4 - Prob. 20RQCh. 4 - Prob. 1PECh. 4 - Prob. 2PECh. 4 - Prob. 3PECh. 4 - Prob. 4PECh. 4 - Prob. 5PECh. 4 - Prob. 6PECh. 4 - Prob. 7PECh. 4 - Prob. 8PECh. 4 - Prob. 9PECh. 4 - Prob. 10PECh. 4 - Prob. 11PECh. 4 - Prob. 1GZCh. 4 - Prob. 2GZ
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