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摩托羅拉moto筆試題
通訊考的都是基礎(chǔ)知識，但是不是通訊專業(yè)的一般都要靠摸的，計算機考的比較有深度，主要是c和c++;類的繼承關(guān)系(初始化順序);二*樹的先序、中序、后序，知道其中兩個，推導(dǎo)剩下的一個;i++和++i問題，
摩托羅拉moto筆試題
。
例如:
int i=1;
int j;
j=(++i)+(++i)+(++i)+(++i);//j=15
int i=1;
int j;
j=(++i)+(++i)+(i++)+(i++)+(++i);//j=16
moto面試中英語口語很重要，技術(shù)面試一開始就是用英語的，后來不行了，只好用中文。第二面的那個人力資源的主管就是全英文了，一路面下來不吐一個漢字。二面的時候主要就是聊天了，問問你的愛好，和同學(xué)關(guān)系什么的，主要是看口語，以及你個人的性格。
moto待遇6000+770。干滿三年一次性發(fā)6000*20%的住房公積金
//計算字符串長度的函數(shù)
int strlen_mod(char* str)
{ int count = 0;
while(str[count] != ‘\0′)
{ ++count;}
return count;
}
//將字符串反轉(zhuǎn)輸出的函數(shù)
void print_reverse(char* str)
{ size_t size = strlen(str);
int size2= strlen_mod(str);
printf(“The length of %s is %d ### %d\n”, str, size, size2);
int i;
char temp = ‘ ‘;
for(i=0; i < size/2; i++)
{ printf(“%d\n”, i);
temp = str[i];
str[i] = str[size - 1 - i];
str[size - 1 - i]= temp;
}
printf(“Reverse string: %s\n”, str);
}
What will print out? 輸出結(jié)果是?
main()
{ char *p1=“name”;
char *p2;
p2=(char*)malloc(20);
memset (p2, 0, 20); //
while(*p2++ = *p1++);
printf(“%sn”,p2);
}
Answer:empty string.
What will be printed as the result of the operation below:
main()
{ int x=20,y=35;
x=y++ + x++;
y= ++y + ++x;
printf(“%d%dn”,x,y);
}
Answer : 5794
What will be printed as the result of the operation below:
main()
{
int x=5;
printf(“%d,%d,%dn”,x,x< <2,x>>2);
}
Answer: 5,20,1
What will be printed as the result of the operation below:
#define swap(a,b) a=a+b;b=a-b;a=a-b;
void main()
{
int x=5, y=10;
swap (x,y);
printf(“%d %dn”,x,y);
swap2(x,y);
printf(“%d %dn”,x,y);
}
int swap2(int a, int b)
{
int temp;
temp=a;
b=a;
a=temp;
return 0;
}
Answer: 10, 5
10, 5
What will be printed as the result of the operation below:
main()
{
char *ptr = ” Cisco Systems”;
*ptr++; printf(“%sn”,ptr);
ptr++;
printf(“%sn”,ptr);
}
Answer:Cisco Systems
isco systems
What will be printed as the result of the operation below:
main()
{
char s1[]=“Cisco”;
char s2[]= “systems”;
printf(“%s”,s1);
}
Answer: Cisco
What will be printed as the result of the operation below:
main()
{
char *p1;
char *p2;
p1=(char *)malloc(25);
p2=(char *)malloc(25);
strcpy(p1,”Cisco”);
strcpy(p2,“systems”);
strcat(p1,p2);
printf(“%s”,p1);
}
Answer: Ciscosystems
The following variable is available in file1.c, who can access it?:
static int average;
Answer: all the functions in the file1.c can access the variable.
WHat will be the result of the following code?
#define TRUE 0 // some code
while(TRUE)
{
// some code
}
Answer: This will not go into the loop as TRUE is defined as 0.
What will be printed as the result of the operation below:
int x;
int modifyvalue()
{
return(x+=10);
}
int changevalue(int x)
{
return(x+=1);
}
void main()
{
int x=10;
x++;
changevalue(x);
x++;
modifyvalue();
printf(“First output:%dn”,x);
x++;
changevalue(x);
printf(“Second output:%dn”,x);
modifyvalue();
printf(“Third output:%dn”,x);
}
Answer: 12 , 13 , 13
What will be printed as the result of the operation below:
main()
{
int x=10, y=15;
x = x++;
y = ++y;
printf(“%d %dn”,x,y);
}
Answer: 11, 16
What will be printed as the result of the operation below:
main()
{
int a=0;
if(a==0)
printf(“Cisco Systemsn”);
printf(“Cisco Systemsn”);
}
Answer: Two lines with “Cisco Systems” will be printed.
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