C_8练习题_代码007(未授权)

本文介绍: 在每小题给出的四个备选项中选出一个正确的答案,并将所选项前的字母填写在答题纸的相应位置上。5,在下面的do-wh ile循环语句中,其循环体语句被执行的次数为()int i=0;4,在sw i t c h 语句的每个 ca se块中,假定都是以b re a k语句结束的,则此sw i t c h语句容易被改写为（）语句。,则下列表达式中值不为2的是(）20,凡是函数中未指定存储类别的局部变量,其隐含的存储类别为(）6,将两个字符串连接起来组成一个字符串时,选用的函数为()10,设x和y均为逻辑假,则xly为假的条件是(.），

一、单项选择题(本大题共20小题,每小题2分,共40分。在每小题给出的四个备选项中选出一个正确的答案,并将所选项前的字母填写在答题纸的相应位置上。)
1,在每个C语言程序中都必须包含有这样一个函数,该函数的函数名为(）

3,设有定义: in t a=5,b; ,则下列表达式中值不为2的是(）

4,在sw i t c h语句的每个ca se块中,假定都是以bre a k语句结束的,则此sw i t c h语句容易被改写为（）语句

5,在下面的do-while循环语句中,其循环体语句被执行的次数为()in t i=0; do i++; while(i<10);

6,将两个字符串连接起来组成一个字符串时,选用的函数为()

#include <std io.h>
void main()
{
int i,x[3][3]={1,2,3,4,5,6,7,8,9};
for(i=0;i<3;i++)
printf(“%d “,x[i][2-i]);,
}

#include<stdio.h&gt;

int main() {
    int a[9] = {21, 25, 56, 24, 39, 62, 18, 87, 64};
    int i, b1, b2;

    a[0] <= a[1] ? (b1 = a[0], b2 = a[1]) : (b1 = a[1], b2 = a[0]);

    for (i = 2; i < 9; i++) {
        if (a[i] < b1) {
            b2 = b1;
            b1 = a[i];
        } else if (a[i] < b2) {
            b2 = a[i];
        }
    }

    printf("%d %dn", b1, b2);

    return 0;
}

#include <stdio.h&gt;

int acc(int n) {
    static int f;
    f = f + n;
    return f;
}

int main() {
    int i;

    for (i = 1; i <= 5; i++) {
        printf("%d ", acc(i));
    }

    return 0;
}

#include <stdio.h&gt;

void SelectSort(int a[], int n) {
    int i, j, k, x;
    for (i = 0; i < n - 1; i++) {
        k = i;
        for (j = i + 1; j < n; j++) {
            if (a[j] < a[k]) {
                k = j;
            }
        }
        x = a[i];
        a[i] = a[k];
        a[k] = x;
    }
}

int main() {
    int i;
    int a[6] = {20, 12, 22, 10, 56, 28};

    SelectSort(a, 6);

    for (i = 0; i < 6; i++) {
        printf("%d ", a[i]);
    }

    printf("n");

    return 0;
}

#include <stdio.h&gt;

void LE(int* a, int* b) {
    int x = *a;
    *a = *b;
    *b = x;
}

int main() {
    int x = 15, y = 26;

    printf("%d %dn", x, y);
    LE(&amp;x, &amp;y);
    printf("%d %dn", x, y);

    return 0;
}

#include <stdio.h&gt;
#include <string.h>

struct Worker {
    char name[15];  // 姓名
    int age;        // 年龄
    float pay;      // 工资
};

int Equal(struct Worker* r1, struct Worker* r2) {
    return strcmp(r1->name, r2->name) == 0;
}

int main() {
    struct Worker a[4] = {
        {"abe", 25, 420},
        {"defa", 58, 638},
        {"ghin", 49, 560},
        {"jklt", 36, 375}
    };

    struct Worker x = {"defa", 0, 0};  // 初始化 x 结构体

    int i;
    for (i = 0; i < 4; i++) {
        if (Equal(a + i, &amp;x)) {
            break;
        }
    }

    if (i == 4) {
        printf("没有查找到所需的记录!n");
    } else {
        printf("%s %d %6.2fn", a[i].name, a[i].age, a[i].pay);
    }

    return 0;
}


#include <stdio.h>

int main() {
    int j, k, sum1 = 0, sum2 = 0;
    int a[4][4];

    printf("Enter an array:n");

    for (j = 0; j < 4; j++) {
        for (k = 0; k < 4; k++) {
            scanf("%d", &amp;a[j][k]);
        }
    }

    sum1 = 0;
    for (j = 0; j < 4; j++) {
        for (___________) {
            sum1 += a[j][k];
        }
    }
    printf("sum1 = %dn", sum1);

    ________;
    for (j = 0; j < 4; j++) {
        for (___________) {
            sum2 += a[j][k];
        }
    }
    printf("sum2 = %dn", sum2);

    return 0;
}

#include <stdio.h>

int compare(char *p1, char *p2) {
    while (*p1++ == *p2++) {
        if (__________) {
            return __________;
  
}

int main() {
    int m;
    char str1[20], str2[20], *p1, *p2;

    printf("Please input strings one by one:n");
    scanf("%s", str1);
    scanf("%s", str2);

    p1 = str1;
    p2 = str2;

    m = compare(p1, p2);

    printf("The result is: %dn", m);

    return 0;
}

#include <stdio.h>
#include <string.h>

int main() {
    char a[6][10] = {"bag", "good", "This", "are", "Zoo", "park"};
    ________;
    int i;

    max = a[0];

    for (i = 1; i < 6; i++) {
        if (strcmp(a[i], max) < 0) {
            max = a[i];
        }
    }

    printf("The min string is %sn", max);

    return 0;
}