作业帮void main() {union { int k; char i[2]; }*s,a; s=&a; s->i[0]=0x39;s->i[1]=0x38; printf("%d\n",s->k);void main(){union{int k;char i[2];}*s,a;s=&a;s->i[0]=0x39;s->i[1]=0x38;printf("%d\n",s->k);}求输出结果（详细解及思路）
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/26 14:44:45
![void main() {union { int k; char i[2]; }*s,a; s=&a; s->i[0]=0x39;s->i[1]=0x38; printf(](/uploads/image/z/3990150-54-0.jpg?t=void+main%28%29+%7Bunion+%7B+int+k%3B+char+i%5B2%5D%3B+%7D%2As%2Ca%3B+s%3D%26a%3B+s-%3Ei%5B0%5D%3D0x39%3Bs-%3Ei%5B1%5D%3D0x38%3B+printf%28%22%25d%5Cn%22%2Cs-%3Ek%29%3Bvoid+main%28%29%7Bunion%7Bint+k%3Bchar+i%5B2%5D%3B%7D%2As%2Ca%3Bs%3D%26a%3Bs-%3Ei%5B0%5D%3D0x39%3Bs-%3Ei%5B1%5D%3D0x38%3Bprintf%28%22%25d%5Cn%22%2Cs-%3Ek%29%3B%7D%E6%B1%82%E8%BE%93%E5%87%BA%E7%BB%93%E6%9E%9C%EF%BC%88%E8%AF%A6%E7%BB%86%E8%A7%A3%E5%8F%8A%E6%80%9D%E8%B7%AF%EF%BC%89)
k);void main(){union{int k;char i[2];}*s,a;s=&a;s->i[0]=0x39;s->i[1]=0x38;printf("%d\n",s->k);}求输出结果（详细解及思路）" />void main() {union { int k; char i[2]; }*s,a; s=&a; s->i[0]=0x39;s->i[1]=0x38; printf("%d\n",s->k);void main(){union{int k;char i[2];}*s,a;s=&a;s->i[0]=0x39;s->i[1]=0x38;printf("%d\n",s->k);}求输出结果（详细解及思路）
void main() {union { int k; char i[2]; }*s,a; s=&a; s->i[0]=0x39;s->i[1]=0x38; printf("%d\n",s->k);
void main()
{union
{
int k;
char i[2];
}*s,a;
s=&a;
s->i[0]=0x39;s->i[1]=0x38;
printf("%d\n",s->k);
}
求输出结果（详细解及思路）
void main() {union { int k; char i[2]; }*s,a; s=&a; s->i[0]=0x39;s->i[1]=0x38; printf("%d\n",s->k);void main(){union{int k;char i[2];}*s,a;s=&a;s->i[0]=0x39;s->i[1]=0x38;printf("%d\n",s->k);}求输出结果（详细解及思路）
union {// 着是个共用体,就是说int的k与char*2的i[2]共用同一个存储空间
int k;
char i[2];
}*s,a;
s=&a;// s指向a的地址
s->i[0]=0x39;// 将16进制的39（2进制00110101）写入a.i[0];
s->i[1]=0x38;// 将16进制的38（2进制00110100）写入a.i[1];
printf("%d\n",s->k);// 输出a.k（2进制0011010100110100）并换行
对了,这里还要看int占几位,
若int占2位则结果是14393;(0x3839)
若int占4位结果就是-859031495;(0xcccc3839)