C语言 将小写字母转换为大写字母

Convert small letter to capital letter

提问by yEL155

I'm trying to convert string in small letter to capital letter. I got some error (access volation) what whould do?

我正在尝试将小写字符串转换为大写字母。我遇到了一些错误（访问波动）怎么办？

int main()
{
    char str[10];
    int i=0;
    scanf("%s", &str);
    while (str[i] !=0)
    {
        str[i] += -32;
        printf("%s", str[i]);
    }
    return 0;
}

thx

谢谢

回答by Sven Marnach

1. If you enter a string longer than 9 characters, scanf()will try to write past the end of your string buffer.

2. Your while-loop never terminates as you never change i.

3. You should use "%c"as format string in your printf()call, since you are wrting characters, not null-terminated strings.

1. 如果输入的字符串长度超过 9 个字符，scanf()将尝试写入超过字符串缓冲区的末尾。

2. 你的 while 循环永远不会终止，因为你永远不会改变i。

3. 您应该"%c"在printf()调用中使用作为格式字符串，因为您正在编写字符，而不是以空字符结尾的字符串。

回答by CyberDem0n

int main()
{
    char str[10];
    int i=0;
    scanf("%s", str);
    while (str[i] != 0)
    {
        str[i] += -32;
            i++;
    }
    printf("%s", str);
    return 0;
}

and of course, you must check overflow of str...

当然，您必须检查 str 的溢出...

回答by apacay

As cprogrammer said

正如程序员所说

You better use toupper

/* toupper example */
#include <stdio.h>
#include <ctype.h>
int main ()
{
  int i=0;
  char str[]="Test String.\n";
  char c;
  while (str[i])
  {
    c=str[i];
    putchar (toupper(c));
    i++;
  }
  return 0;
}

你最好使用toupper

/* toupper example */
#include <stdio.h>
#include <ctype.h>
int main ()
{
  int i=0;
  char str[]="Test String.\n";
  char c;
  while (str[i])
  {
    c=str[i];
    putchar (toupper(c));
    i++;
  }
  return 0;
}

But if not, and you want to do it your way

但如果没有，而且你想按照自己的方式去做

int main()
{
    char str[10];
    int i=0;
    scanf("%s", &str);
    while (str[i]!='
#include<stdio.h>
#include<conio.h>
#include<string.h>

void main();
{
    int i, count;
    char str[200];
    clrscr();
    printf("Enter a string");
    scanf("%s", str);
    count = strlen(str);
    for(i=0; i<=count; i++)
    {
        if((str[i] >= 97) && (str[i] <= 122))
        {
            str[i] = str[i] - 32;
        }
    }
    printf("%s", str);
    getch();
}
' && i<10)  
    {// You forgot this: '
char ch = letter & 223; [letter = a-z] // Now ch is all time capital letter
' instead of 0 and also i<10
        str[i] += -32;
        printf("%c", str[i]);//char, not string
        i++; //And this
    }
    return 0;
}

回答by MByD

There are few mistakes here:

这里有几个错误：

1. scanf("%s", &str);- since str is a pointer to char, you don't need to give its address, but scanf("%s", str);. (and as sven said, it's unsafe)
2. while (str[i] !=0)this is an endless loop, you should increment iat the end of the while block.

3. str[i] += -32;will modify any char you're at, you should check if this is a lower case any time, for example:

   if (str[i] >= 'a' && str[i] <= 'z'){ str[i] -=32; } //couldn't format this line for some reason....

4. printf("%s", str[i])is again wrong way to use printf, since %sexpects to char*, and str[i]is a char. instead, use printf("%c", str[i])which expects a char

1. scanf("%s", &str);- 因为 str 是一个指向 char 的指针，所以你不需要给出它的地址，但是scanf("%s", str);. （正如 sven 所说，这是不安全的）
2. while (str[i] !=0)这是一个无限循环，您应该i在 while 块的末尾增加。

3. str[i] += -32;将修改您所在的任何字符，您应该随时检查这是否是小写，例如：

   if (str[i] >= 'a' && str[i] <= 'z'){ str[i] -=32; } //由于某种原因无法格式化此行....

4. printf("%s", str[i])又是使用 printf 的错误方式，因为%s期望char*，并且str[i]是char. 相反，使用printf("%c", str[i])它需要一个字符

回答by Aneesh

/* toupper example */
#include <stdio.h>
#include <ctype.h>
int main ()
{
  int i=0;
  char str[]="Test String.\n";
  char c;
  while (str[i])
  {
    c=str[i];
    putchar (toupper(c));
    i++;
  }
  return 0;
}

回答by Arif

You can use Bitwise AND( &) operator technique to make small letter to capital letter.

您可以使用按位 AND( &) 运算符技术将小写字母转换为大写字母。

##代码##

回答by cprogrammer

You better use toupper

你最好使用toupper

##代码##