c - 内存寻址和指针练习

Question

假设您有一台 16 位机器，内存为 65536 字节。整数为 2 个字节。假设变量按照编码顺序连续放置在内存中，从内存地址 0x250 开始。给定以下代码段:

int x = 30; 
int y = 50; 
int *px = &x;
int *py = &y;

printf("a) %x\n", px); 
printf("b) %x\n", py); 

px = py; 
printf("c) %d\n", *px); 
printf("d) %x\n", &px); 

x = 88; 
y = 14;

printf("e) %d\n", *px); 
printf("f) %d\n", *py); 

给出每个 printf 语句的输出。

他们说假设内存地址是按升序分配的。

我得到

a) 252
b) 254
c) 30
d) 250
e) 88
f) 14

这些正确吗？

Answer 1

Assume that variables are placed in memory contiguously in the order they are coded, starting at memory address 0x250

让我们遵循 nneonneo 的提示，“作为第一步，只需分配变量地址，这样您就知道所有内容都在哪里。”

这应该让你迈出正确的一步:

int x = 30;               // &x  = 0x250
int y = 50;               // &y  = 0x252
int *px = &x;             // &px = 0x254
int *py = &y;             // &py = 0x256

现在有一点指导:

printf("a) %x\n", px);    // These both print the *value* of the pointers 
printf("b) %x\n", py);    // themselves (in other words, the address to which
                          // they point).
                          // Note: they're printed in hexadecimal
                          // (unfortunately with no prefix).

px = py;                  // The *value* of py is assigned to px.
                          // Now they both point to the same thing.


printf("c) %d\n", *px);   // This "de-references" px. In other words, it
                          // reads the value at the address px points to.


printf("d) %x\n", &px);   // This prints the *address* of the pointer px.
                          // Hint: can this change during program execution?

x = 88;                   // Simple integer assignments
y = 14;

printf("e) %d\n", *px);   // Same logic as c)

printf("f) %d\n", *py);   // Same logic as c)

答案(对于那些想检查答案的人或作弊者):

a) 250
b) 252
c) 50
d) 254
e) 14
f) 14