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#include "stdio.h" int main() { char c_buf[5] = {1,2,3,4,5}; int z_buf[5] =
{6,7,8,9,10}; char *c = c_buf; int *z = z_buf; printf("sizeof(c) =
%ld\n",sizeof(c)); printf("sizeof(z) = %ld\n",sizeof(z)); printf("c_add =
%p\n",c); printf("z_add = %p\n",z); printf("c_add+1 = %p\n",c+1);
printf("z_add+1 = %p\n",z+1); return 0; } 运行结果如下： sizeof(c) = 8 sizeof(z) = 8
c_add = 0x7ffea5a70960 z_add = 0x7ffea5a70940 c_add+1 = 0x7ffea5a70961 z_add+1
= 0x7ffea5a70944
分析下代码和答案，我们就可以发现以下几点：
1、指针占用空间大小一致，均为8字节，因为在64位系统上执行，故一个地址占8个字节。
2、char *和int *自增的偏移量不同，char*偏移一位，int*偏移四位。
故结论为char*和int*得差别在于内存自增的偏移大小
