#includevoid main(){int aa[4][4]={{1,2,3,4},{5,6,7,8},{3,9,10,2},{4,2,9,6}};#includevoid main(){int aa[4][4]={{1,2,3,4},{5,6,7,8},{3,9,10,2},{4,2,9,6}};int i,s=o;for(i=o;i,4;i++)s+=aa[i][2];printf("%d\n",s);的结果是什么
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 22:55:18
![#includevoid main(){int aa[4][4]={{1,2,3,4},{5,6,7,8},{3,9,10,2},{4,2,9,6}};#includevoid main(){int aa[4][4]={{1,2,3,4},{5,6,7,8},{3,9,10,2},{4,2,9,6}};int i,s=o;for(i=o;i,4;i++)s+=aa[i][2];printf(](/uploads/image/z/2733445-37-5.jpg?t=%23includevoid+main%28%29%7Bint+aa%5B4%5D%5B4%5D%3D%7B%7B1%2C2%2C3%2C4%7D%2C%7B5%2C6%2C7%2C8%7D%2C%7B3%2C9%2C10%2C2%7D%2C%7B4%2C2%2C9%2C6%7D%7D%3B%23includevoid+main%28%29%7Bint+aa%5B4%5D%5B4%5D%3D%7B%7B1%2C2%2C3%2C4%7D%2C%7B5%2C6%2C7%2C8%7D%2C%7B3%2C9%2C10%2C2%7D%2C%7B4%2C2%2C9%2C6%7D%7D%3Bint+i%2Cs%3Do%3Bfor%28i%3Do%3Bi%2C4%3Bi%2B%2B%29s%2B%3Daa%5Bi%5D%5B2%5D%3Bprintf%28%22%25d%5Cn%22%2Cs%29%3B%E7%9A%84%E7%BB%93%E6%9E%9C%E6%98%AF%E4%BB%80%E4%B9%88)
#includevoid main(){int aa[4][4]={{1,2,3,4},{5,6,7,8},{3,9,10,2},{4,2,9,6}};#includevoid main(){int aa[4][4]={{1,2,3,4},{5,6,7,8},{3,9,10,2},{4,2,9,6}};int i,s=o;for(i=o;i,4;i++)s+=aa[i][2];printf("%d\n",s);的结果是什么
#includevoid main(){int aa[4][4]={{1,2,3,4},{5,6,7,8},{3,9,10,2},{4,2,9,6}};
#include
void main()
{int aa[4][4]={{1,2,3,4},{5,6,7,8},{3,9,10,2},{4,2,9,6}};
int i,s=o;
for(i=o;i,4;i++)s+=aa[i][2];
printf("%d\n",s);
的结果是什么
#includevoid main(){int aa[4][4]={{1,2,3,4},{5,6,7,8},{3,9,10,2},{4,2,9,6}};#includevoid main(){int aa[4][4]={{1,2,3,4},{5,6,7,8},{3,9,10,2},{4,2,9,6}};int i,s=o;for(i=o;i,4;i++)s+=aa[i][2];printf("%d\n",s);的结果是什么
第一次运行i=2,sum[&aa[2]]就把aa[2]当做首地址传给函数sum,然后sum函数中的a[0],a[1]分别代表了aa[2],aa[3],所以第一次完毕以后aa[2]=aa[3]=4;然而aa[0]并没有改变
然后第二次运行i=1;同样将4传给了aa[1]=4;
然后第三次i=0,将aa[0]赋值为4.所以结果应该是4
希望对你能有所帮助.