无法获取 arr[] : Implicit conversion loses integer precision: 'unsigned long' to 'int' for strlen(arr) 的长度

Question

隐式转换会丢失整数精度:
'unsigned long' 到 'int'

我想获取单词的长度[]但它告诉我这个。我该如何解决？

int main(int argc, const char * argv[]) {
    
    char test[] = "ls test";
    f_sparce_arg(test);
    return 0;
}
char * f_sparce_arg(char words[]){
    int words_length;
    words_length = strlen(words);   /* warning: Implicit conversion 
                                       integer precision: 'unsigned 
                                       long' to 'int' */
    
    printf("%d", words_length);
    return "";
}

// warning: Implicit conversion integer precision: 'unsigned long' to 'int'

Answer 1

什么 strlen function返回的是 size_t，所以你应该用它来存储它的返回值。

char * f_sparce_arg(char words[]){
    size_t words_length; /* use proper type */
    words_length = strlen(words);

    printf("%zu",words_length); /* the format to print size_t value is %zu */
    return "";
}

不幸的是，某些编译器或库不支持 %zu 格式。另一种方法是，如果字符串不会太长，则将返回值转换为 int。

char * f_sparce_arg(char words[]){
    int words_length;
    words_length = (int)strlen(words); /* explicitly cast the result to avoid warnings */

    printf("%d",words_length);
    return "";
}