C 指针管理(地址、取消引用、重新分配)
C Pointer Mangement (Address, Dereferencing, Reassigning)
我正在尝试编写一个函数来访问一个字节并将其更改为另一个字节。我在管理指针时遇到问题。
谁能帮我修改这段代码,让 px 指向地址,然后我可以递增 pc(这与 px 相同,但就像 char 一样,以避免递增太多字节),然后替换原来的内容在 mem 位置 pc 和其他东西?
unsigned replace_byte(unsigned x, int i, unsigned char b){
unsigned int* px = &x; //Takes initial address of first byte.
unsigned char* pc= (unsigned char*)px; //Recast it too char
//pointer to avoid incrementing by too many bytes
if (i==0) {
*pc= b; //if i is 0 then replace the byte at mem location pc
//with b
}
if (i==1) { //if i is 1 or more then inc by that much and replace
pc = pc+1;
*pc= b;
}
if (i==2) {
pc=pc+2;
*pc= b;
}
if (i==3) {
pc=pc+3;
*pc= b;
}
return x;
}
我得到的 return 值如下:
305419947
305441656
313218680
2872333944
当我想获得这样的值时:
replace_byte(0x12345678, 0xAB, 2) --> 0x12AB5678
replace_byte(0x12345678, 0xab, 0) --> 0x123456AB
这对我来说非常有效:
#include <stdio.h>
unsigned replace_byte(unsigned x, int i, unsigned char b){
unsigned int px = x;
unsigned char* pc= (unsigned char*) &px;
if (i > 3) return px;
*(pc + i) = b;
return px;
}
int main()
{
int a = 0x12345678;
a = replace_byte(a, 2, 0x7F); // any hexadecimal character in parameter 3
printf("%x", a);
}
此代码演示如何更改 int 类型的特定字节。
#include <stdio.h>
int main() {
unsigned int i = 0xffffffff;
//// make two pointers to the one memory address
// x - pointer to int. The address will be shifting by 4 bytes, when this
// pointer will be shifting by 1.
int *x = &i;
// y - pointer to char. The address will be shifting by 1 byte, when this
// pointer will be shifting by 1.
char *y = (char *) x;
//addresses the same here.
printf("x address = %p\n", x);
printf("y address = %p\n", y);
puts("");
//both pointers are shifting by 1, but addresses are different now.
printf("x + 1 = %p\n", x+1);
printf("y + 1 = %p\n", y+1);
puts("");
//changing the 'i' original value by one byte in turns, one after another.
printf("i - original: %x\n", i);
*y = 16;
printf("i - the first byte changed: %x\n", i);
*(y+1) = 16;
printf("i - the second byte changed: %x\n", i);
*(y+2) = 16;
printf("i - the third byte changed: %x\n", i);
*(y+3) = 16;
printf("i - the forth byte changed: %x\n", i);
return 0;
}
输出:
x address = 0x7fffe59e9794
y address = 0x7fffe59e9794
x + 1 = 0x7fffe59e9798
y + 1 = 0x7fffe59e9795
i - original: ffffffff
i - the first byte changed: ffffff10
i - the second byte changed: ffff1010
i - the third byte changed: ff101010
i - the forth byte changed: 10101010
我正在尝试编写一个函数来访问一个字节并将其更改为另一个字节。我在管理指针时遇到问题。
谁能帮我修改这段代码,让 px 指向地址,然后我可以递增 pc(这与 px 相同,但就像 char 一样,以避免递增太多字节),然后替换原来的内容在 mem 位置 pc 和其他东西?
unsigned replace_byte(unsigned x, int i, unsigned char b){
unsigned int* px = &x; //Takes initial address of first byte.
unsigned char* pc= (unsigned char*)px; //Recast it too char
//pointer to avoid incrementing by too many bytes
if (i==0) {
*pc= b; //if i is 0 then replace the byte at mem location pc
//with b
}
if (i==1) { //if i is 1 or more then inc by that much and replace
pc = pc+1;
*pc= b;
}
if (i==2) {
pc=pc+2;
*pc= b;
}
if (i==3) {
pc=pc+3;
*pc= b;
}
return x;
}
我得到的 return 值如下: 305419947 305441656 313218680 2872333944
当我想获得这样的值时:
replace_byte(0x12345678, 0xAB, 2) --> 0x12AB5678 replace_byte(0x12345678, 0xab, 0) --> 0x123456AB
这对我来说非常有效:
#include <stdio.h>
unsigned replace_byte(unsigned x, int i, unsigned char b){
unsigned int px = x;
unsigned char* pc= (unsigned char*) &px;
if (i > 3) return px;
*(pc + i) = b;
return px;
}
int main()
{
int a = 0x12345678;
a = replace_byte(a, 2, 0x7F); // any hexadecimal character in parameter 3
printf("%x", a);
}
此代码演示如何更改 int 类型的特定字节。
#include <stdio.h>
int main() {
unsigned int i = 0xffffffff;
//// make two pointers to the one memory address
// x - pointer to int. The address will be shifting by 4 bytes, when this
// pointer will be shifting by 1.
int *x = &i;
// y - pointer to char. The address will be shifting by 1 byte, when this
// pointer will be shifting by 1.
char *y = (char *) x;
//addresses the same here.
printf("x address = %p\n", x);
printf("y address = %p\n", y);
puts("");
//both pointers are shifting by 1, but addresses are different now.
printf("x + 1 = %p\n", x+1);
printf("y + 1 = %p\n", y+1);
puts("");
//changing the 'i' original value by one byte in turns, one after another.
printf("i - original: %x\n", i);
*y = 16;
printf("i - the first byte changed: %x\n", i);
*(y+1) = 16;
printf("i - the second byte changed: %x\n", i);
*(y+2) = 16;
printf("i - the third byte changed: %x\n", i);
*(y+3) = 16;
printf("i - the forth byte changed: %x\n", i);
return 0;
}
输出:
x address = 0x7fffe59e9794
y address = 0x7fffe59e9794
x + 1 = 0x7fffe59e9798
y + 1 = 0x7fffe59e9795
i - original: ffffffff
i - the first byte changed: ffffff10
i - the second byte changed: ffff1010
i - the third byte changed: ff101010
i - the forth byte changed: 10101010