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This topic has been don't understand, o great god answer

CodePudding user response:

Type A, B, sure is correct, and the answer?

CodePudding user response:

reference 1/f, the truth is more important than right or wrong response:

A, B, sure is correct, and the answer?

1 is the answer I don't understand why the array name as a function parameter, executive function, why s value does not change? Is it not return s?

CodePudding user response:

Pass in a pointer to the function, change in a function pointer, after the function pointer remains unchanged, but the internal change function pointer to the address, after the function that address will change, the content of the

CodePudding user response:

reference 3 floor truth is right or wrong response:

to a pointer to the function, the change in a function pointer, after the function pointer remains unchanged, but the function of the internal pointer to change address, after the function that address will change, the content of the

is the function of address? Or s address

CodePudding user response:

Use the malloc function in the s new out, so after operation is not the first incoming s

CodePudding user response:

M0_46236484

reference 4 floor response:

Quote: refer to the third floor truth is right or wrong response:
pass in a pointer to the function, change in a function pointer, after the function pointer remains unchanged, but the function of the internal pointer to change address, after the function that address will change, the content of the

is the function of address? Or s address

What is the relationship between function's address?

Such as void test (int a) {
A=10;
}
So you so call
Int x=20;
The test (x);
//here or 20 x; Not be 10;
Pointer, a pointer is the address, the address is a number, said a memory area,
So the change in the function pointer, outside have no effect, unless the address change the content of the area of memory,

CodePudding user response:

The result is a [0] o value is 1.
The reason is that s=(int *) malloc (sizeof (int)); This line to s new address assignment, so fun function will not have any effect,

CodePudding user response:

refer to 6th floor truth is right or wrong response:

Quote: refer to 4th floor m0_46236484 response:

Quote: refer to the third floor truth is right or wrong response:
to a pointer to the function, the change in a function pointer, after the function pointer remains unchanged, but the function of the internal pointer to change address, after the function that address will change, the content of the

is the function of address? Or s address

What is the relationship between function's address?

Such as void test (int a) {
A=10;
}
So you so call
Int x=20;
The test (x);
//here or 20 x; Not be 10;
Pointer, a pointer is the address, the address is a number, said a memory area,
So the change in the function pointer, outside have no effect, unless you change the address of the content of the memory area,

I don't know much about this zha analysis

CodePudding user response:

reference 7 floor cold wind response: 1023

the result is a [0] o value is 1.
The reason is that s=(int *) malloc (sizeof (int)); This line to s new address assignment, so fun function will not have any effect,

yes, yes, I understand

CodePudding user response:

If there is no C, D option, there is no answer, the answer should be 1.

Because the parameter is the value of transmission, the s and s in the main func is independent of each other two Pointers, of operation does not affect the main func on s s,