代码片段为：
char c[]="gate2011";
char *p=c;
printf("%s",p+p[3]-p[1]);
The output is: 2011
有人能解释一下它是怎么来的吗？
-----提前感谢-----
依次通过每一行：
char c[] = "gate2011";
让我们假设阵列c位于存储器地址200处。
char *p = c;
p现在是指向c的指针。因此它指向存储器地址200。p的实际内容是“200”，表示内存地址。
printf("%s", p + p[3] - p[1]);
当我们把p当作指针对待时，它的值是200。然而，我们也可以把它当作一个数组。p[3]获取字符串中第4项的值，即“e”。C将字符存储为ASCII值。“e”的ASCII值为101。
接下来，我们得到p[1]的值。p[1]==“a”，ASCII值为97。将这些替换为函数：
printf("%s", 200 + 101 - 97);
其评估结果为：
printf("%s", 204);
在存储器地址204处，我们有字符串“2011”。因此，程序打印“2011”。
我不知道你为什么要做这样的事情，但无论如何，这就是正在发生的事情。
p + p[3] - p[1]
这里取一个指针的值，将位置3处的char值相加，然后减去位置1处的char的值。在进行加法和减法运算之前，字符值被隐式转换为数值。
如果是1 000个地点,则将支付1 000+101(e)-97(a)美元。 因此,结果是发现1004个记忆器。 <>>印本/>中的百分数则以该地点开始的体格为准,并以特殊性质终止。 因此,扼杀实际上被拖到“2011年”(头4封信由于101-97=4而丢失)。
如果这仍然没有意义，我建议您仔细研究一下C中的数组是如何工作的。
你期待什么？为什么不呢？
p[3]-p[1] = e - a = 4
p+4 = "2011"
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