和利時(shí)筆試題

連接兩個(gè)有序列表head1和head2；（從小到大的順序）typedefstructstu{intnum;structstu*next;}STU;STU*link_chain(STU*head1,STU*head2){STU*head,*p,*p1,*p2;p1=head1;p2=head2;if((p1->num)<=(p2->num)){p=head=p1;p1=p1->next;}else{p=head=p2;p2=p2->next;}while((p1!=NULL)||(p2!=NULL)){if(p1==NULL){p->next=p2;returnhead;}if(p2==NULL){p->next=p1;returnhead;}if((p1->num)<=(p2->num)){p->next=p1;p=p1;p1=p1->next;}else{p->next=p2;p=p2;p2=p2->next;}}}======================================================================和利時(shí)======================================================================intmain(){intj=2;inti=1;if(i=1)j=3;if(i=2)j=5;printf("%d",j);}宏定義是在哪個(gè)階段被首先處理？預(yù)編譯階段。編程題：(一）1位轉(zhuǎn)換位87654321數(shù)v8v7v6v5v4v3v2v1轉(zhuǎn)換后：位87654321數(shù)v1v2v3v4v5v6v7v8執(zhí)行越快越好，可以用空間代價(jià)換取時(shí)間。答：unsignedcharbit_reverse(unsignedcharc){unsignedcharbuf=0;intbit=8;while(bit){bit--;buf|=((c&1)<<bit);c>>=1;}returnbuf;}（二）結(jié)構(gòu)體structcont2{char*str1;char*str2;};實(shí)現(xiàn)函數(shù)structcont2*Newcont2(char*s1,char*s2);函數(shù)功能為，先申請一個(gè)結(jié)構(gòu)體變量的空間，再為結(jié)構(gòu)體中變量申請空間，若申請失敗，要釋放已經(jīng)申請的所有空間?？梢允褂脦旌瘮?shù)，strlen(),strcpy(),malloc().答：structcont2*Newcont2(char*s1,char*s2){structcont2*buf=NULL;buf=(structcont2*)malloc(sizeof(structcont2));if(buf==NULL)returnNULL;buf->str1=(char*)malloc(strlen(s1)+1);if(buf->str1==NULL){free(buf);returnNULL;}strcpy(buf->str1,s1);buf->str2=(char*)malloc(strlen(s2)+1)if(buf->str2==NULL){free(buf->str1);free(buf);returnNULL;}strcpy(buf->str2,s2);returnbuf;}(三）實(shí)現(xiàn)intswap（int*a，int*b），不使用中間變量。答：intswap(int*a,int*b){*a=*a+*b;*b=*a-*b;*a=*a-*b;}（四）設(shè)計(jì)一個(gè)命令