unsigned short A = 10; printf("~A = %u\n",A); char c=128; printf("c=%d\n",c); 输出多少?第一题,A =0xfffffff5,int值 为-11,但输出的是uint.所以输出4294967285 第二题,c=0x10,输出的是int,最高位为1,是负数,所以它的值

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unsigned short A = 10; printf(

unsigned short A = 10; printf("~A = %u\n",A); char c=128; printf("c=%d\n",c); 输出多少?第一题,A =0xfffffff5,int值 为-11,但输出的是uint.所以输出4294967285 第二题,c=0x10,输出的是int,最高位为1,是负数,所以它的值
unsigned short A = 10; printf("~A = %u\n",A); char c=128; printf("c=%d\n",c); 输出多少?
第一题,A =0xfffffff5,int值 为-11,但输出的是uint.所以输出4294967285 第二题,c=0x10,输出的是int,最高位为1,是负数,所以它的值就是0x00的补码就是128,所以输出-128.这两道题都是在考察二进制向int或uint转换时的最高位处理.A =0xfffffff5,还有为什么c=0x10,不应该是c=0x08吗,不好意思各位大侠,小弟的分数用完了.

unsigned short A = 10; printf("~A = %u\n",A); char c=128; printf("c=%d\n",c); 输出多少?第一题,A =0xfffffff5,int值 为-11,但输出的是uint.所以输出4294967285 第二题,c=0x10,输出的是int,最高位为1,是负数,所以它的值
2^32=4294967296,
A=10,为无符号型,转换为二进制为0000 0000 0000 0000 0000 0000 0000 1010
所以~A的二进制位1111 1111 1111 1111 1111 1111 1111 0101即0xFFFFFFF5,如果转换为符号整型的话则为-11,因为输出的是无符号整型,无符号整型的范围为0~4294967295,而0xFFFFFFF5转换为无符号十进制整型为4294967285
第二题,发生溢出,因为有符号字符型其范围为-128~127
127用二进制表示为:0111 1111,128表示为1000 0000,这里发生溢出,因为第一位为1,为符号位,表示负数,即-128

unsigned short us =0xABCD,us 求补码怎么求的...#include #include void main (){unsigned short int a;unsigned short int getbits(unsigned short);printf ( input an octal number:);scanf (%o,&a);printf (%o ,getbits(a));}unsigned short int getbits (unsigned short value){u 希望给出注释~unsigned short GetHashValue(unsigned int *grid){unsigned int mask1 = 524287; //19个1unsigned int mask2 = 8191; //13个1unsigned short len = 65535;unsigned int result =0;result += ((grid[0] & (mask113)|((grid[0] & mask2) unsigned short a=6,b=3; 一、表达式a=b,求表达式的值. c语言表达式求值 unsigned short x=1 shrot v 问表达式的值v=~x C语言中 unsigned short 怎么输出?例如 int d ;printf(d = %d ,d);用'%d' ,那么unsigned short 当给 16 位的 unsigned short 对象赋值 100 000 时,赋的值是什么?100000-65535-1=34464 unsigned char ctoa(char a[]){unsigned char c=0;for(int i=0;i unsigned short A = 10; printf(~A = %u ,A); char c=128; printf(c=%d ,c); 输出多少?第一题,A =0xfffffff5,int值 为-11,但输出的是uint.所以输出4294967285 第二题,c=0x10,输出的是int,最高位为1,是负数,所以它的值 连求什么都不知.一、填空:(1) 以下程序的输出结果是 【1】  main()  { unsigned short a=65536; int b;  printf(“%d ”,b=a) ;  }(2) 若有定义:int a=10,b=9,c=8; 接着顺序执行下列语句后,变量b中的 unsigned int gcd(unsigned int a,unsigned int b){while(b^=a^=b^=a%=b);return a;} 有一个hash函数unsigned short hashf(unsigned short key){ return (key>>4)%256}请问hashf(16)和hashf(256)的结果,会的人做一下阿,有分加的1!最好说一下理由 #include main() {unsigned int a=65535; printf(a=%d ,a); } 运算结果 a=-1#includemain(){unsigned int a=65535;printf(a=%d ,a);}运算结果a=-1为什么是等于-1呢? a short time =( ) unsigned char a[4] = {0,0,0,1};int b = *((int*)a); 这样执行完毕后,b的值是多少?unsigned char a[4] = {0,0,0,1};int b = *((int*)a);这样执行完毕后,b的值为什么不是1,而是16777216,而unsigned char a[4] = {0,0,0,0}; b的值就可 关于位域运算 麻烦帮解释下下列程序#includevoid main(){struct bs{unsigned a:1;unsigned b:3;unsigned c:4;} bit,*pbit;bit.a=1;bit.b=7;bit.c=15;printf(%d,%d,%d ,bit.a,bit.b,bit.c);pbit=&bit;pbit->a=0; //此后三行是什么意思pbit-> 谁能告诉我以下程序结果,unsigned fun6(unsigned num){unsigned k=1;do{k*=num%10;num/10;}while(num);return(k);}main(){unsigned n=26;printf(%d ,fun6(n));}是c语言题 为何unsigned short是65535数值长度,short是32767数值长度