下面的C程序片段打印了什么?
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char c[] = "GATE2011" ; char *p =c; printf ( "%s" , p + p[3] - p[1]) ; |
(A) 盖茨2011 (B) E2011 (C) 2011 (D) 011 答复: (C) 说明: 有关解释,请参见注释。
char c[] = "GATE2011"; // p now has the base address string "GATE2011" char *p = c; // p[3] is 'E' and p[1] is 'A'. // p[3] - p[1] = ASCII value of 'E' - ASCII value of 'A' = 4 // So the expression p + p[3] - p[1] becomes p + 4 which is // base address of string "2011" printf("%s", p + p[3] - p[1]); // prints 2011
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