如何在不使用数组表示法的情况下将字符串从小写转换为大写? [英] How do I convert a string from lowercase to uppercase without using array notation?
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问题描述
任务是将字符串从小写转换为大写而不使用数组表示法?
我给出的提示是修改字符串指针本身,我以为你无法修改字符串指针。我不知道如何处理while循环。
任何帮助都会很棒。
我尝试过:
The task is to convert a string from lowercase to uppercase without using array notation?
The hint I was given was "modify the string pointer itself", I thought you couldn't modify a string pointer. I'm not sure how to work the while loop.
Any help would be great.
What I have tried:
int main(void)
{
uppercase_no_array("reverse");
return(0);
}
void uppercase_no_array(char* input)
{
char* copy = malloc(sizeof(char) * strlen(input));
strcpy(copy, input);
printf("%s", copy);
while(copy != '\0')
{
}
return;
}
推荐答案
分配内存时,复制
指向空间中的第一个字符。然后使用它将输入复制到输出区域 - 根本不需要这样做,只需使用已有的两个指针:copy
和输入
When you allocate the Memory,copy
points at the first character in the space. You then use that to copy the input to an output area - you don't really need to do that at all, just use the two pointers you already have:copy
andinput
char* uppercase_no_array(char* input)
{
char* mem = malloc(sizeof(char) * (strlen(input) + 1));
char* copy = mem;
char c;
do
{
c = *input++;
... uppercase it here ...
*copy++ = c;
} while(c != '\0')
return mem;
}
Quote:
我给出的提示是修改字符串指针本身
The hint I was given was "modify the string pointer itself"
从我的角度来看,这意味着你可以使用指针就地修改字符串:
From my point of view that means that you can modify the string in-place using a pointer:
char *p;
p = input_string;
while (*p)
{
/* char_to_upper() has to be implemented */
*p = char_to_upper(*p);
p++;
}
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