c - char 字符串中的 ascii 代码

Question

我如何从数据生成器获取 datapoint*25+65 是像 a 这样的单个字符，但我想在 str = abcdefg 中包含(哪个字母无关紧要) )? datapoint 创建一个介于 0.0 和 1.0 之间的值。单字符程序:

char str;
for(int n; n<60; n++)
{
    str=datapoint*25+65;
    str++;
}
str = '/0';

问题是我不知道如何使用此设置获得像 abcd 这样的字符字符串，而不仅仅是像 str 中的 a 这样的单个字母。

Answer 1

试试这个:

char str[60 + 1]; /* Make target LARGE enough. */
char * p = str; /* Get at pointer to the target's 1st element. */
for(int n = 0; /* INITIALISE counter. */ 
    n<60; 
    n++)
{
  *p = datapoint*25+65; /* Store value by DE-referencing the pointer before
                         assigning the value to where it points. */
  p++; /* Increment pointer to point to next element in target. */
}
*p = '\0'; /* Apply `0`-terminator using octal notation, 
              mind the angle of the slash! */

puts(str); /* Print the result to the console, note that it might (partly) be
              unprintable, depending on the value of datapoint. */

没有指向当前元素但使用索引的替代方法:

char str[60 + 1]; /* Make target LARGE enough. */
for(int n = 0; /* INITIALISE counter. */ 
    n<60; 
    n++)
{
  str[n] = datapoint*25+65; /* Store value to the n-th element. */
}
str[n] = '\0'; /* Apply `0`-terminator using octal notation, 
                  mind the angle of the slash! */