作业帮25. 下列程序运行后的输出结果是（）.main() { char a[10]={'1','2','3','4','5','6','7','8','9',0},*p; int i=8; p=a+i; printf("%s\n",p-3);}A.123456789 B.6789 C.'6' D.789求解释过程...
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![25. 下列程序运行后的输出结果是（）.main() { char a[10]={'1','2','3','4','5','6','7','8','9',0},*p; int i=8; p=a+i; printf(](/uploads/image/z/13606371-27-1.jpg?t=25.+%E4%B8%8B%E5%88%97%E7%A8%8B%E5%BA%8F%E8%BF%90%E8%A1%8C%E5%90%8E%E7%9A%84%E8%BE%93%E5%87%BA%E7%BB%93%E6%9E%9C%E6%98%AF%EF%BC%88%EF%BC%89.main%28%29++++%7B+char+a%5B10%5D%3D%7B%271%27%2C%272%27%2C%273%27%2C%274%27%2C%275%27%2C%276%27%2C%277%27%2C%278%27%2C%279%27%2C0%7D%2C%2Ap%3B++++++int+i%3D8%3B++++++p%3Da%2Bi%3B++++++printf%28%22%25s%5Cn%22%2Cp-3%29%3B%7DA.123456789+++++++B.6789++++++++C.%276%27++++++D.789%E6%B1%82%E8%A7%A3%E9%87%8A%E8%BF%87%E7%A8%8B...)
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25. 下列程序运行后的输出结果是（）.main() { char a[10]={'1','2','3','4','5','6','7','8','9',0},*p; int i=8; p=a+i; printf("%s\n",p-3);}A.123456789 B.6789 C.'6' D.789求解释过程...
25. 下列程序运行后的输出结果是（）.
main()
{ char a[10]={'1','2','3','4','5','6','7','8','9',0},*p;
int i=8;
p=a+i;
printf("%s\n",p-3);
}
A.123456789 B.6789 C.'6' D.789
求解释过程...
25. 下列程序运行后的输出结果是（）.main() { char a[10]={'1','2','3','4','5','6','7','8','9',0},*p; int i=8; p=a+i; printf("%s\n",p-3);}A.123456789 B.6789 C.'6' D.789求解释过程...
选B
“1234567890”相当于一个字符转,a为该字符串的首地址,即指向‘1’内存单元的地址,
a+8,为内存单元‘9’所对应的地址,
即p=a+8-3
所以p指向的是a+5内存单元（即‘6’）,所以从‘6’开始输出字符串,遇到‘\0’(即0)则停止.
所以为6789
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