【C/C++--ERROR】：scanf_s函数获取数组时需要2个参数（名称，字符长度）

1.代码：talkback.c – 演示与用户交互

// talkback.c -- 演示与用户交互
#include <stdio.h>
#include <string.h>  // 提供strlen()函数的原型
#define DENSITY 62.4 // 人体密度（单位：磅/立方英尺）
int main(void) {
    
    
	float weight, volume;
	int size, letters;
	char name[40];  //name是一个可容纳40个字符的数组

	printf("Hi, What's your first name?\n");
	scanf_s("%s", name);
	printf("%s,what's your weight in pounds?\n", name);
	
	scanf_s("%f", &weight);
	size = sizeof name;
	letters = strlen(name);
	volume = weight / DENSITY;
	printf("well,%s,your volume is %2.2f cubic feet.\n", name, volume);
	printf("also,your first name has %d letters,\n", letters);
	printf("and we have %d bytes to store it.\n", size);

	return 0;
}

2.报错如下：

在编译时会出现如下警告，且运行时scanf_s后续代码无法运行：

1>e:\talkback.c(17): warning C4244: “=”: 从“double”转换到“float”，可能丢失数据
1>e:\talkback.c(11): warning C4473: “scanf_s”: 没有为格式字符串传递足够的参数
1>e:\talkback.c(11): note: 占位符和其参数预计 2 可变参数，但提供的却是 1 参数
1>e:\talkback.c(11): note: 缺失的可变参数 2 为格式字符串“%s”所需
1>e:\talkback.c(11): note: 此参数用作缓冲区大小

3.原因分析：

根据上面警告提示，scanf_s的调用还缺少一个参数（没有传递足够的参数），scanf_s是scanf的安全版本，需要表示数组长度的参数，所以正确的调用应该是：

scanf_s("%s", name, 40);

4.完整代码：

// talkback.c -- 演示与用户交互
#include <stdio.h>
#include <string.h>  // 提供strlen()函数的原型
#define DENSITY 62.4 // 人体密度（单位：磅/立方英尺）
int main(void) {
    
    
	float weight, volume;
	int size, letters;
	char name[40];  //name是一个可容纳40个字符的数组

	printf("Hi, What's your first name?\n");
	scanf_s("%s", name, 40);
	printf("%s,what's your weight in pounds?\n", name);
	
	scanf_s("%f", &weight);
	size = sizeof name;
	letters = strlen(name);
	volume = weight / DENSITY;
	printf("well,%s,your volume is %2.2f cubic feet.\n", name, volume);
	printf("also,your first name has %d letters,\n", letters);
	printf("and we have %d bytes to store it.\n", size);

	return 0;
}