129 lines
2.2 KiB
Markdown
129 lines
2.2 KiB
Markdown
---
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title: 2025-2026学年春季期期中(张泉)
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---
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## 一、判断（16 分）
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1. 变量定义可以数字开头
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2. C 语言是面向过程的语言
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3. `(x % 2 == 0) && (6 <= x <= 10)` 表示 $[6,10]$ 的偶数
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4. 不符合逻辑顺序的代码如果代码紧凑，则符合书写规范(大意如此)
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5. 假定 `a, b` 是 `double` 类型。`printf("%lf", (a = 2, b = a + 3 / 2))` 输出是 `3.000000`
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## 二、读代码（25 分）
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1. a=\_\_\_\_
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```c
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#include<stdio.h>
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int main() {
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int a = 3, b = 4;
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a += b *= a + b;
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printf("a=%d", a);
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}
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```
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2. a=\_\_\_\_<br></br>a=\_\_\_\_
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```c
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#include<stdio.h>
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void Anaesthesia(int b) {
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static int a = 0;
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a += b;
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printf("a=%d\n", a);
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}
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int main() {
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Anaesthesia(5);
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Anaesthesia(5);
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}
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```
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3. \_\_\_\_, \_\_\_\_
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```c
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#include<stdio.h>
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int main() {
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c = 'A' + 4 - 'Z' + 'z';
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d = 'A' + '8' - '5';
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printf("%c %d", c, d);
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}
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```
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4. \_\_\_\_, \_\_\_\_
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```c
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#include<stdio.h>
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int main() {
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int a = 2;
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char b;
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double Pi = 3.14;
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printf("%d %lf", sizeof(a*b), Pi+8/3);
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}
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```
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5. \_\_\_\_
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```c
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#include<stdio.h>
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int main() {
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int a = 5, b = 10, c = 15;
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if((a -= 5) || (b = a) && (c -= 1)) {
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c += 5;
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} else {
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c -= 5;
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}
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printf("%d", c);
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}
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```
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## 三、选择题（25分）（注：题面暂缺）
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函数不允许嵌套定义函数（函数里不可以定义函数）
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编译成功一定运行一定成功
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变量定义是下划线、字母、数字随意组合
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3. 正确的是（）
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A. `char` 可以取余
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B. `(a = 3, b = 5, a + b)` 的值是 $3$
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C. `i++` 和 `++i` 任何情况下都一样
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D. 赋值运算符优先级高于所有基本运算符
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## 四、手写代码（35 分）
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1. 定义一个常量 $E=2.71828$， 输出 $E^3$ （保留五位小数）
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**Input**
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> 无
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**Output**
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> `20.08553`
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2. 给入一个值 `n`， 打出表格
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**Input**
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> `4`
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**Output**
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> ```text
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> ***#
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> **#*
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> *#**
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> #***
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> ```
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3. 输入一个字母，小写转大写，大写转小写
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**Input**
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> `a`
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**Output**
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> `A`
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