题目内容

#includemain() { char c1,c2; c1='A'+'8'-'4'; c2='A'+'8'-'5'; printf("%c,%d\n",c1,c2); } It is known that the ASCII code of letter A is 65, and the output of the program is ( )

A. E,68
B. D,69
C. E,D
D. Output has no fixed value

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#includemain () {int a=1,b=2; while (a<6) {b+=a; a+=2; b%=10;} printf("%d,%d\n",a,b); } the output of the program is ( )

A. 5，11
B. 7，1
C. 7，11
D. 6，1

There are the following procedures： #include int main(void) { int a=1,b=2,m=0,n=0,k; k=(n=b>a)||(m=a

A. 0,0
B. 0,1
C. 1,0
D. 1,1

The illegal expression in the following options is ( )

A. 0<=x<100
B. i=j==0
C. (char)(65+3)
D. x+1=x+1

If the variables are correctly defined and assigned, the following legal C language assignment statement is（）.

A. x=y==5;
B. x=n%2.5;
C. x+n=i;
D. x=5=4+1;