{

//Exchange the highest byte and the lowest byte

int n 1 = 0x ff 000000 & amp； n；

int n2 = 0x000000ff & ampn；

n =(n & amp； 0x 00 ffffff)|(N2 & lt； & lt24);

n =(n & amp； 0x ffffff 00)| n 1；

//Swap the middle two bytes

n 1 = 0x 00ff 0000 & amp； n；

n2 = 0x0000ff00 & ampn；

n =(n & amp； 0x ff 00 ffff)|(N2 & lt； & lt8);

n =(n & amp； 0x ffff 00 ff)|(n 1 & gt； & gt8);

Returns n;

}

int main()

{

int n = 12345678；

printf( "n=0x%x\n "，n)；

int N2 = RevertHex(n)；

printf( "n2=0x%x\n "，N2)；

Returns 0;

}

The hexadecimal code of 12345678 is: 0xBC6 14E (it can also be regarded as 0x 0bc 6 14E). After conversion, it is 0x4e6 1bc00.

If you don't want it at 00, you can think of another way.