The c language-CodePudding

#include

Void free (int I)
{
If (I.=0)
i=0;
}

Int main ()
{
Int I, [5] a={1, 2, 3, 4, 5};
Free (a [2]);
I=a, [2].
Printf (" % d \ n ", I);
}

The result is 3, why is not zero? Parameters and arguments of knowledge?

CodePudding user response:

As parameter arguments of knowledge (parameter is a copy of the argument), involving preach value and the address of the knowledge

CodePudding user response:

Because I was free in the function of the local variable, rather than global or variables in the main function, the value of the local variable cannot I=0 to the main function,

CodePudding user response:

1/f, said makes sense, it is a question of value to pass or address, you can change
Void free (int * p)
{
If (* p!=0) {
* p=0;
}
}

CodePudding user response:

reference 2 building self-confidence boy reply:

because I was free in the function of the local variable, rather than global or variables in the main function, the value of the local variable cannot I=0 to the main function,

If this is the case then?
#include

Void free (int m)
{
If (m!=0)
M=0;
}

Int main ()
{
Int I, [5] a={1, 2, 3, 4, 5};
Free (a [2]);
I=a, [2].
Printf (" % d \ n ", I);
}

CodePudding user response:

refer to Hans - 123-4 floor response:

Quote: refer to the second floor confident boy reply:
of a local variable in a function because I was free, rather than global or variables in the main function, the value of the local variable cannot I=0 to the main function,

If this is the case then?
#include

Void free (int m)
{
If (m!=0)
M=0;
}

Int main ()
{
Int I, [5] a={1, 2, 3, 4, 5};
Free (a [2]);
I=a, [2].
Printf (" % d \ n ", I);
}

The m is not in the same, free of m in the main, the parameter m in the free of a local variable in a function is free, but you can do it:

 void free (int * m) 
{
If (* m!=0) 
* m=0; 
}

This definition free, namely the argument addresses to free,

CodePudding user response:

https://blog.csdn.net/yinzewen123/article/details/80532886? Ops_request_misc=% 257 b % 2522 request 2522% % 255 fid % 253 a 2522% % 2522% % 2522159920883319725222423450% % 2522 SCM 252 c 253 a % 252220140713.130102334 PC % 255 fall. The 2522% % 257 d & amp; Request_id=159920883319725222423450 & amp; Biz_id=0 & amp; Utm_medium=distribute. Pc_search_result. None - task - blog - 2 ~ all ~ first_rank_ecpm_v3 ~ pc_rank_v3-3-80532886. Pc_ecpm_v3_pc_rank_v3 & amp; E5 utm_term=% 95% % 87% % % E6 BD B0 E8 8 d B8 E4 % % % % % 83% % BD E8 E5 % BF % 94% % 9 b % 9 A0 e % E6 % % 88% E5 E9 8 f % 98% % 87% % E7 8 f % % a % 84% % E6 8 of 9 E9 c % 87% % 92% 88 & amp; SPM=1018.2118.3001.4187
You can take a look at this link, the most basic should be the cause of the stack memory

CodePudding user response:

Void free (int I)
{
If (I.=0)
i=0;
} function calls out, variable I statement cycle will be over, because I was stack memory variables, so after use is unknown,

CodePudding user response:

In c + +, reference can be used directly, and others don't move,,,

Void free (int & amp; M)
{
If (m!=0)
M=0;
}