Objective-C隐式转换会失去整数精度(将size_t转换为CC_Long) [英] Objective-C Implicit conversion loses integer precision (size_t to CC_Long)

问题描述

我有一个函数可以生成字符串的sha256加密，

I have a function that's generating a sha256 encryption of a string,

功能如下:

    -(NSString*)sha256HashFor:(NSString*)input
{
    const char* str = [input UTF8String];
    unsigned char result[CC_SHA256_DIGEST_LENGTH];
    CC_SHA256(str, strlen(str), result);

    NSMutableString *ret = [NSMutableString stringWithCapacity:CC_SHA256_DIGEST_LENGTH*2];
    for(int i = 0; i<CC_SHA256_DIGEST_LENGTH; i++)
    {
        [ret appendFormat:@"%02x",result[i]];
    }
    return ret;
}

现在此行在这里CC_SHA256(str, strlen(str), result);是产生此警告的内容(该警告是针对strlen(str)变量的.)

Now this line right here CC_SHA256(str, strlen(str), result); is what's producing this warning (the warning is for the strlen(str) variable).

Implicit conversion loses integer precision: 'size_t' (aka 'unsigned long') to 'CC_LONG' (aka 'unsigned int')

我猜我只需要将strlen(str)转换为CC_Long，但是我不知道该怎么做.

I'm guessing I just need to convert the strlen(str) to a CC_Long, but I have no idea how to do that.

推荐答案

1. 大概不是错误，而是警告.

1. Presumably that's not an error but a warning.

我只需要将strlen(str)转换为CC_Long，但是我不知道该怎么做." -显式类型转换(类型转换):(CC_LONG)strlen(str)，但我认为您并不是真的需要它.

"I just need to convert the strlen(str) to a CC_Long, but I have no idea how to do that." - explicit type conversion (type casting): (CC_LONG)strlen(str), but I don't think you really need this.

这篇关于Objective-C隐式转换会失去整数精度(将size_t转换为CC_Long)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！