重大復(fù)試基礎(chǔ)測試題

1、Cpp試題1（A-I 1*11=11分解決1、voidFunc ( char str =Cpp試題1（A-I 1*11=11分解決1、voidFunc ( char str =char n =str = ;(str =( p ) = ( n ) = void *p = malloc( sizeof ( p ) = 3、Struct Charb; Chard; FloatSizeof(tag)= 4、Struct CharShortc:5; Char e:2; Char d;5;Char 6Charx0 = 輸出為： 7、CSwitch 3 答：char ;Char 6Charx0 = 輸出為：

2、 7、CSwitch 3 答：char 類型，case “&ra=a,。9.將13關(guān)于“聯(lián)合”include union char voida.x0 =a.x1= #defineDOUBLE(x)x+x voida.x0 =a.x1= #defineDOUBLE(x)x+x ，i5*DOUBLE(5)i 是多少main 描述內(nèi)存分配方式以及它們的區(qū)別17.18.const與#define 3Constdefine19.1、char *GetMemory(void)char p = oworld; return p;void char *str = NULL; str = 請(qǐng)問運(yùn)行Test2、V

3、oidGetMemory2(char*p = (char void char *str = NULL; GetMemory(&str,100);strcpy(str, Test3voidGetMemory(charp = (char void char*strstrcpy(str, Test3voidGetMemory(charp = (char void char*str=NULL; strcpy(str, o 請(qǐng)問運(yùn)行Test答4void char*str=(char strcpy(str, “ if(str != *) strcpy(str, Test21改錯(cuò)題試題1void char

4、char*str1=strcpy(string,str1試題2void charstring11,char*str1=strcpy(string,str1試題2void charstring11,str111; for(i=0;i10;str1i=strcpy(string,str1試題3voidtest3(char*char if(strlen(str1)=10strcpy(string,str1試題4*p2) *p= *p1=*p2=countx=0; countxx=x&(x-countx=0; countxx=x&(x-returncountx;x9999structdata ; No

5、de*next;typedefstructNodeNode24strcpy 函數(shù)（10strcpy*strcpy(char*strDest,constchar*strSrc); strDest 是目的字符串，strSrc 是源字符串。（1）不調(diào)用C+/C25、strncpymemcpy26、#ifndefdefine#endif27new、delete、malloc、free3033、當(dāng)一個(gè)類A 中沒有任何成員變量與成員函數(shù),這時(shí)sizeof(A)34、String 的具體實(shí)現(xiàn)class Stringpublic:String(constchar*strNULL); / String(constString&a