main() { int a=2;a%=4-1; printf("\n%d,",a); a+=a*=a-=a*=3; printf("%d",a); }
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main(){inta=2;a%=4-1;printf("\n%d,",a);a+=a*=a-=a*=3;printf("%d",a);}main(){inta=2;a%=4-1;printf("\n
main() { int a=2;a%=4-1; printf("\n%d,",a); a+=a*=a-=a*=3; printf("%d",a); }
main() { int a=2;a%=4-1; printf("\n%d,",a); a+=a*=a-=a*=3; printf("%d",a); }
main() { int a=2;a%=4-1; printf("\n%d,",a); a+=a*=a-=a*=3; printf("%d",a); }
int a=2;
a%=3;即a=2%3=2
第一个输出2
后面一串,从右到左
a-=a*=3就已经是0了.
所以最后一直都是0
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37、有以下程序 int a=2; int f(int *a) {return (*a)++;} main( ) { int s=0; { int a=5; s+=f(&a); } s+答案为什么是C,37、有以下程序 int a=2; int f(int *a) {return (*a)++;} main( ) { int s=0; { int a=5; s+=f(&a); } s+=f(&a); printf(%d
void fun(int a,int b) { int t; t=a;a=b;b=t; } main()void fun(int a,int b){ int t; t=a;a=b;b=t; } main() { int c[10]={1,2,3,4,5,6,7,8,9,0}.i; for(i=0;i
#include using namespace std; int main() { int a,b,c; a=3; int f(int x,int y,int z);#include using namespace std; int main() {int a,b,c;a=3;int f(int x,int y,int z); cin>>a>>b>>c;c=f(a,b,c);cout
int a=2; int f(int a); {return (a)++;} main() {int s=0; {int a=5; s+=f(&&a);} s+=f(&&a); printf(%da=2;int f(int a);{return (a)++;}main(){int s=0;{int a=5;s+=f(&&a);}s+=f(&&a);printf(%d
,s);}执行的输出结果是()A 10 B 9 C 7 D 8