first question
1. Among the following options, the operator that cannot operate on pointer variables of the same basic type is ()
A：+
B：-
C：=
D：==
Answer and analysis A
A: Error, pointers cannot be added, because pointer addition may go out of bounds, so it is not allowed;
B: Correct, pointer subtraction is used to find the number of data of the same data type inside.
C: Pointers are allowed to be assigned
D: Determine the equality of pointers, that is, if the addresses are different, if the addresses are the same, the variables inside are also the same;
Question 2
2. The output result of the following program segment is ( )
int a, b;
b = (a = 3*5, a*4, a*5);
printf("%d",b);A：60
B：75
C：65
D: No definite value
Answer and analysis B
This question examines a comma expression. The comma expression calculates the formulas between the commas in sequence, but the result of the comma expression is the result of the last formula inside;
So the results of (a = 3*5, a * 4, a * 5) are (1, 60, 75) respectively
Remember that a has been assigned a value in the first expression, so subsequent a will have this value.
The final answer is 75
Question 3
3. The following expression has the same result as ++*p ()
int a[]= {1,2,3,4,5};
int *p = a;A：*++p
B：a[0]
C：a[0]++
D：*p++
Answer and analysis A
This question tests the usage of priorities and pointers
Although the priority of prefix ++ is higher than *, the associativity of prefix ++ is from right to left. The value of the left operand must be calculated first and cannot exceed *, so *p = 1 is calculated first; Then prepend ++, and finally ++*p = 2;
Related blogs:C language operator priority table (recommended to collect and read it every time) - CSDN Blog
A: *++p, the precedence of ++ is higher than *, and the precedence of ++ returns the value after ++, so *++p is equivalent to *(p + 1) = 2;
B：a[ 0 ] = 1
C: a[ 0 ]++, followed by ++, returns the value before ++, so the value of a[ 0 ]++ expression is 1;
D: *p++, postfixed ++ has a higher priority than *, but postfixed ++ returns the current value, so the address of p in this expression is still a, so it is equivalent to *p, *p = * a = 1;
Question 4
4. int *p[4] is equivalent to () in the selection
A：int p[4]
B：int *p
C：int *(p[4])
D：int (*p)[4]
Answer and analysis C
First of all, the [ ] of int *p [4] in the question has a priority higher than *
So p is first combined with [4] to form an array. Each element in the array is of type int *, which is a pointer array;
A: int p[4], integer array;
B: int *p, integer pointer;
C: int * (p[4]) is enclosed in parentheses and is an array of pointers;
D: int (*p) [4] This is a pointer, pointing to an array, and is an array pointer;
Question 5 (detailed review!!!!!)
5. Assume that the function prototype and variable description are as follows, and the illegal call is ()
void f3(int(*p)[4]);
int a[4]={1，2，3，4}，
int b[3][4]={
{1，2，3，4}，{5，6，7，8}，{9，10，11，12}};A：f3(&a);
B：f3(b[1]);
C：f3(&b[1]);
D：f3(b);
Answer and analysis B
In fact, when a function passes parameters, the type must match; this question tests those array pointers;
Because int (*p)[4] is a pointer to an array of pointers; and how can a pointer point to the entire array? It must be the address of the stored array, which means that the type of p is actually &array name, &a. p is equivalent to a secondary pointer which is int**, but the difference is that p must also have the number of arrays. match;
A: &a, a is the array name, &a represents the address of the entire array, so it matches the parameter type of f3;
B: b[1], b is a two-dimensional array. The row name of the two-dimensional array is equivalent to the array name of each row. Because the two-dimensional array is actually an array of one-dimensional arrays, then each row is equivalent to a One-dimensional array, isn't the row name the array name of our one-dimensional array, so b[1] is the array name, the array name is the address of the first element, and the first element of the row b[1] is b[1][0 ], the address type is int*, which does not match our p type.
C: &b[1] is correct, it is the address of the entire array;
D: b is the array name, the name of the two-dimensional array, which is the address of the first element of the two-dimensional array. The first element of the two-dimensional array is the entire one-dimensional array, so the data type of b is actually the address of the one-dimensional array, and p type matching; because if you think about it, the first element of a one-dimensional array is a single number, but a two-dimensional array is an array of one-dimensional arrays, so the elements of the two-dimensional array must be one-dimensional arrays one by one!
