在Xcode中寻址C程序

概述当我在 Xcode中的64位Intel中编译下面的代码时,我得到了这个输出. #include<stdio.h>#include<limits.h>int main(void){ /* declare some integer variables */ long a = LONG_MAX; long b = 2L; long c = 3L; /* declare so 当我在 Xcode中的64位Intel中编译下面的代码时,我得到了这个输出.

#include<stdio.h>#include<limits.h>int main(voID){  /* declare some integer variables */  long a = LONG_MAX;  long b = 2L;  long c = 3L;  /* declare some floating-point variables */  double d = 4.0;  double e = 5.0;  double f = 6.0;  printf("A variable of type long occupIEs %d bytes.",sizeof(long));  printf("\nHere are the addresses of some variables of type long:");  printf("\nThe address of a is: %p  The address of b is: %p",&a,&b);  printf("\nThe address of c is: %p",&c);  printf("\nThe address of a-b is: %ld\nvalue of a is %ld\nValue of b is %ld\nsize of pointer %ld ",(&a-&c),a,b,sizeof(&a));  printf("\n\nA variable of type double occupIEs %d bytes.",sizeof(double));  printf("\nHere are the addresses of some variables of type double:");  printf("\nThe address of d is: %p  The address of e is: %p",&d,&e);  printf("\nThe address of f is: %p\n",&f);    printf("\n size long - %d",sizeof(a));  return 0;}

A variable of type long occupIEs 8 bytes.Here are the addresses of some variables of type long:The address of a is: 0x7fff5fbff880 The address of b is: 0x7fff5fbff878 The address of c is: 0x7fff5fbff870 The address of a-b is: 2value of a is 9223372036854775807 Value of b is 2 size of pointer 8 A variable of type double occupIEs 8 bytes.Here are the addresses of some variables of type double:The address of d is: 0x7fff5fbff868 The address of e is: 0x7fff5fbff860 The address of f is: 0x7fff5fbff858 size long - 8

对我来说奇怪的是a和b的地址之间的差异只有2.我希望它是8,这将与long的字节数相匹配.有谁知道为什么会这样？

我在代码中输入了一个拼写错误& a-& c,但这真的与我的问题无关.我的问题是,为什么从变量a的地址到变量b的地址只有2个字节的差异,当long为8个字节长时,我希望看到8的差异？

解决方法 差异以sizeof(long)为单位.要强制它以字节为单位,你应该先抛出两个指针：

((char *)&a-(char *)&b)

以这种方式,差异以sizeof(char)为单位,即字节数

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