int a[ ]={1,2,3,4,5,6,7,88,9,10},*p=a; 则值为3的表式是( ) A) p+=2,*(p++) B) p+=2,*++p C) p+=3,*pint a[ ]={1,2,3,4,5,6,7,88,9,10},*p=a;则值为3的表式是( )A) p+=2,*(p++) B) p+=2,*++pC) p+=3,*p++ D) p+=2,++*p选什么,为什么?
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![int a[ ]={1,2,3,4,5,6,7,88,9,10},*p=a; 则值为3的表式是( ) A) p+=2,*(p++) B) p+=2,*++p C) p+=3,*pint a[ ]={1,2,3,4,5,6,7,88,9,10},*p=a;则值为3的表式是( )A) p+=2,*(p++) B) p+=2,*++pC) p+=3,*p++ D) p+=2,++*p选什么,为什么?](/uploads/image/z/2737470-30-0.jpg?t=int+a%5B+%5D%3D%7B1%2C2%2C3%2C4%2C5%2C6%2C7%2C88%2C9%2C10%7D%2C%2Ap%3Da%3B+%E5%88%99%E5%80%BC%E4%B8%BA3%E7%9A%84%E8%A1%A8%E5%BC%8F%E6%98%AF%28+%29+A%29+p%2B%3D2%2C%2A%28p%2B%2B%29+B%29+p%2B%3D2%2C%2A%2B%2Bp+C%29+p%2B%3D3%2C%2Apint+a%5B+%5D%3D%7B1%2C2%2C3%2C4%2C5%2C6%2C7%2C88%2C9%2C10%7D%2C%2Ap%3Da%3B%E5%88%99%E5%80%BC%E4%B8%BA3%E7%9A%84%E8%A1%A8%E5%BC%8F%E6%98%AF%28+%29A%29+p%2B%3D2%2C%2A%28p%2B%2B%29+B%29+p%2B%3D2%2C%2A%2B%2BpC%29+p%2B%3D3%2C%2Ap%2B%2B+D%29+p%2B%3D2%2C%2B%2B%2Ap%E9%80%89%E4%BB%80%E4%B9%88%2C%E4%B8%BA%E4%BB%80%E4%B9%88%3F)
int a[ ]={1,2,3,4,5,6,7,88,9,10},*p=a; 则值为3的表式是( ) A) p+=2,*(p++) B) p+=2,*++p C) p+=3,*pint a[ ]={1,2,3,4,5,6,7,88,9,10},*p=a;则值为3的表式是( )A) p+=2,*(p++) B) p+=2,*++pC) p+=3,*p++ D) p+=2,++*p选什么,为什么?
int a[ ]={1,2,3,4,5,6,7,88,9,10},*p=a; 则值为3的表式是( ) A) p+=2,*(p++) B) p+=2,*++p C) p+=3,*p
int a[ ]={1,2,3,4,5,6,7,88,9,10},*p=a;
则值为3的表式是( )
A) p+=2,*(p++) B) p+=2,*++p
C) p+=3,*p++ D) p+=2,++*p
选什么,为什么?
int a[ ]={1,2,3,4,5,6,7,88,9,10},*p=a; 则值为3的表式是( ) A) p+=2,*(p++) B) p+=2,*++p C) p+=3,*pint a[ ]={1,2,3,4,5,6,7,88,9,10},*p=a;则值为3的表式是( )A) p+=2,*(p++) B) p+=2,*++pC) p+=3,*p++ D) p+=2,++*p选什么,为什么?
选A,题中把数组的首地址赋给了P之后,那么*p==1,要让他为3,当然是然他后移两位,即p+=2,此时*p=3,至于后面的*(p++)和*(++p)区别么,p++是先让p输出然后自增1,++p就是先让p自增1然后在输出.当p后移两位时,实际此时*(p)==3,为了让结果为3,只能是先输出在自增了.实际上此题主要是靠你的自增运算的用法,实际上要让输出为3,大可不必有后面的自增运算,只有前面的运算已经足够了!