a = [ i for i in range(1, 8) if i % 2 == 1 ],那么sum(a)的值是( )。
a = [ i for i in range(1, 8) if i % 2 == 1 ],那么sum(a)的值是( )。
下面程序段的输出结果是()。int i;int a[2][4]={2,4,6,8,10,12,14,16}for(i=0;i<2;i++)printf("%d,",a[1-i][i+2]); A: 10,6 B: 14,8 C: 12,4 D: 4,10
下面程序段的输出结果是()。int i;int a[2][4]={2,4,6,8,10,12,14,16}for(i=0;i<2;i++)printf("%d,",a[1-i][i+2]); A: 10,6 B: 14,8 C: 12,4 D: 4,10
Complete the following function: /* copy string2 to string1 */ void strcopy(char string1[], char string2[]) { int i = 0; while (string2[i] != '\0') { string1[i] = string2[i]; i++; } ____<br/>}[/i][/i][/i] A: return; B: string1[i - 1] = '\0'; C: string1[i] = '\0'; D: string1[i + 1] = '\0';
Complete the following function: /* copy string2 to string1 */ void strcopy(char string1[], char string2[]) { int i = 0; while (string2[i] != '\0') { string1[i] = string2[i]; i++; } ____<br/>}[/i][/i][/i] A: return; B: string1[i - 1] = '\0'; C: string1[i] = '\0'; D: string1[i + 1] = '\0';
分析下面程序段的运行结果是( )。 char str1[]="abcd",str2[]="abcef"; int i,s; i=0; while((str1[i]==str2[i])&&(str1[i]!='\0')) i++; s=str1[i]-str2[i]; printf("%d\n",s);[/i][/i][/i][/i][/i] A: -1 B: 0 C: 1 D: 不确定
分析下面程序段的运行结果是( )。 char str1[]="abcd",str2[]="abcef"; int i,s; i=0; while((str1[i]==str2[i])&&(str1[i]!='\0')) i++; s=str1[i]-str2[i]; printf("%d\n",s);[/i][/i][/i][/i][/i] A: -1 B: 0 C: 1 D: 不确定
vara=[];for(vari=0;i<10;i++){a[i]=function(){console.log(i);};}a[6]();输出结果是1。[/i]
vara=[];for(vari=0;i<10;i++){a[i]=function(){console.log(i);};}a[6]();输出结果是1。[/i]
已知 vec=[[1,2],[3,4]],则表达式[[row[i] for row in vec] for i in range(len(vec[0]))]的值为( )[/i]
已知 vec=[[1,2],[3,4]],则表达式[[row[i] for row in vec] for i in range(len(vec[0]))]的值为( )[/i]
已知列表lst=[1,2,3],则分别执行以下代码,结果为[9,4,1]的是: A: for i in lst: i=i**2print(lst) B: for i in range(0,3): lst[i]**=2print(lst) C: for i in range(1,4): lst[3-i]=i**2print(lst) D: for i in range(2,-1,-1): lst[i]**=2print(lst)
已知列表lst=[1,2,3],则分别执行以下代码,结果为[9,4,1]的是: A: for i in lst: i=i**2print(lst) B: for i in range(0,3): lst[i]**=2print(lst) C: for i in range(1,4): lst[3-i]=i**2print(lst) D: for i in range(2,-1,-1): lst[i]**=2print(lst)
下面代码运行结果是()。publicclassTest{publicstaticvoidmain(String[]args){int[]a=newint[3];int[]b=newint[]{1,2,3,4,5};a=b;for(inti=0;i<=b.length;i++){System.out.print(a[i]+ " "); }}}[/i] A: 1 2 3 4 B: 1 2 3 C: 0 1 2 3 4 D: 数组越界异常
下面代码运行结果是()。publicclassTest{publicstaticvoidmain(String[]args){int[]a=newint[3];int[]b=newint[]{1,2,3,4,5};a=b;for(inti=0;i<=b.length;i++){System.out.print(a[i]+ " "); }}}[/i] A: 1 2 3 4 B: 1 2 3 C: 0 1 2 3 4 D: 数组越界异常
已知 vec = [[1,2], [3,4]],则表达式 [[row[i] for row in vec] for i in range(len(vec[0]))]的值为______________()_________。[/i] A: [1, 2, 3, 4] B: [[1, 2, 3], 4] C: [[1, 3], [2, 4]] D: [1, 2, [3, 4]]
已知 vec = [[1,2], [3,4]],则表达式 [[row[i] for row in vec] for i in range(len(vec[0]))]的值为______________()_________。[/i] A: [1, 2, 3, 4] B: [[1, 2, 3], 4] C: [[1, 3], [2, 4]] D: [1, 2, [3, 4]]
下面语句执行结果为( ) public static void main(String[] args){ int i=4; if(i<0){ i=1;}
下面语句执行结果为( ) public static void main(String[] args){ int i=4; if(i<0){ i=1;}