为什么当我向它传递一个数字后带有无效字符的字符串时，atof() 函数不是 return 0？
Why does the atof() function not return 0 when I pass it a string with invalid characters after a number?
我需要将字符串转换为浮点数。如果字符串不是数字，我希望 return 0.
我已尝试使用以下代码测试 atof() 函数是否适用于此：
printf("%f", atof("1a"));
根据我对atof的理解，当atof不能转换时，ed的值return是0，而这一行却打印出1.0.
为什么会这样？通过文档，我了解到只要输入不是数字，atof 就意味着 return 0。
我看到你查看了一些文档，但你一定错过了强调的部分——确保你彻底阅读文档很重要：http://www.cplusplus.com/reference/cstdlib/atof/
The function first discards as many whitespace characters (as in isspace) as necessary until the first non-whitespace character is found. Then, starting from this character, takes as many characters as possible that are valid following a syntax resembling that of floating point literals (see below), and interprets them as a numerical value. The rest of the string after the last valid character is ignored and has no effect on the behavior of this function.
因此，该函数的行为完全正常 - "1a" 中的 a 字符被忽略。
来自 doc。强调我的
The function first discards as many whitespace characters (as in
isspace) as necessary until the first non-whitespace character is
found. Then, starting from this character, takes as many characters
as possible that are valid following a syntax resembling that of
floating point literals (see below), and interprets them as a
numerical value. The rest of the string after the last valid character
is ignored and has no effect on the behavior of this function.
yet this line prints 1.0.
printf("%f", atof("1a"));
之所以有效，是因为 atof 可以解析至少一个数字，即在这种情况下 1。
但是：
char a[]="a1";
printf("%f\n",atof(a));
应该如您所料给您一个 return 值 0.0。
我需要将字符串转换为浮点数。如果字符串不是数字，我希望 return 0.
我已尝试使用以下代码测试 atof() 函数是否适用于此：
printf("%f", atof("1a"));
根据我对atof的理解，当atof不能转换时，ed的值return是0，而这一行却打印出1.0.
为什么会这样？通过文档，我了解到只要输入不是数字，atof 就意味着 return 0。
我看到你查看了一些文档，但你一定错过了强调的部分——确保你彻底阅读文档很重要：http://www.cplusplus.com/reference/cstdlib/atof/
The function first discards as many whitespace characters (as in isspace) as necessary until the first non-whitespace character is found. Then, starting from this character, takes as many characters as possible that are valid following a syntax resembling that of floating point literals (see below), and interprets them as a numerical value. The rest of the string after the last valid character is ignored and has no effect on the behavior of this function.
因此，该函数的行为完全正常 - "1a" 中的 a 字符被忽略。
来自 doc。强调我的
The function first discards as many whitespace characters (as in isspace) as necessary until the first non-whitespace character is found. Then, starting from this character, takes as many characters as possible that are valid following a syntax resembling that of floating point literals (see below), and interprets them as a numerical value. The rest of the string after the last valid character is ignored and has no effect on the behavior of this function.
yet this line prints 1.0.
printf("%f", atof("1a"));
之所以有效，是因为 atof 可以解析至少一个数字，即在这种情况下 1。
但是：
char a[]="a1";
printf("%f\n",atof(a));
应该如您所料给您一个 return 值 0.0。