如何在C语言中将整数值转换为ASCII字符?我想将字符分配给字符数组。
char buff[10];
假设我们有:
int = 93 (HEX: 5D) -> result should be - buff = {']'}
int = 13398 (HEX: 3456) -> result should be buff = {'4', 'V'}
与here类似
我不需要关心不可打印的字符。总会有可打印的字符。
只需使用位移来获取单个字节。
假设int
的大小为4的架构:
int someInt = ...
uint8_t first = (someInt >> 24);
uint8_t second = (someInt >> 16);
uint8_t third = (someInt >> 8);
uint8_t fourth = someInt;
现在您可以将结果字节放入数组中。一定要检查first
,second
和third
以确保它们不是0
,如果它们是,则跳过它们。确保根据C字符串的要求使用null终止符结束数组。
这个答案假定为big-endian排序,因为这是你在你的例子中所表明的。如果你想要little-endian,只需在将它们放入数组时反转字节的顺序。
请注意,这将把5DC
变成05
和DC
。如果你想要5D
,你应该检查原来的int
中的第一个数字是否是0
。您可以使用&
运算符执行此操作,针对int
,0xf0000000
等测试0x00f00000
。如果您发现第一个数字是0
,请将int
向右移动4位,然后从中提取字节。
所以,像这样:
void ExtractBytes(int anInt, uint8_t *buf, size_t bufSize) {
// passing an empty buffer to this function would be stupid,
// but hey, doesn't hurt to be idiot-proof
if (bufSize == 0) { return; }
// Get our sizes
const int intSize = sizeof(anInt);
const int digitCount = intSize * 2;
// find first non-zero digit
int firstNonZero = -1;
for (int i = 0; i < digitCount; i++) {
if ((anInt & (0xf << ((digitCount - 1 - i) * 4))) != 0) {
firstNonZero = i;
break;
}
}
if (firstNonZero < 0) {
// empty string; just bail out.
buf[0] = 0;
return;
}
// check whether first non-zero digit is even or odd;
// shift if it's odd
int intToUse = (firstNonZero % 2 != 0) ? (anInt >> 4) : anInt;
// now, just extract our bytes to the buffer
int bufPtr = 0;
for (int i = intSize - 1; i >= 0; i--) {
// shift over the appropriate amount, mask against 0xff
uint8_t byte = (intToUse >> (i * 8));
// If the byte is 0, we can just skip it
if (byte == 0) {
continue;
}
// always check to make sure we don't overflow our buffer.
// if we're on the last byte, make it a null terminator and bail.
if (bufPtr == bufSize - 1) {
buf[bufPtr] = 0;
return;
}
// Copy our byte into the buffer
buf[bufPtr++] = byte;
}
// Now, just terminate our string.
// We can be sure that bufPtr will be less than bufSize,
// since we checked for that in the loop. So:
buf[bufPtr] = 0;
// Aaaaaand we're done
}
现在让我们来看看:
uint8_t buf[10];
ExtractBytes(0x41424344, buf, 10);
printf("%s\n", buf);
ExtractBytes(0x4142434, buf, 10);
printf("%s\n", buf);
和输出:
ABCD
ABC
用C语言将整数值转换为ASCII字符?...
参考ASCII table,C中']'
的值将始终被解释为0x5D或十进制值93.而C中“]”的值将始终被解释为NULL
终止的char数组,即包含的字符串表示形式价值:
|93|\0|
(如This Answer所示,类似的解释对所有ASCII字符都有效。)
要将任何整数(char
)值转换为看起来像“]”的值,可以使用字符串函数将char
值转换为字符串表示形式。例如,所有这些变体都将执行该转换:
char strChar[2] = {0};
sprintf(strChar, "%c", ']');
sprintf(strChar, "%c", 0x5D);
sprintf(strChar, "%c", 93);
并且每个产生相同的C字符串:"]"
。
我想将字符分配给字符数组...
如何创建char数组的示例,以NULL
char结尾,例如“ABC ... Z”:
int i;
char strArray[27] = {0};
for(i=0;i<26;i++)
{
strArray[i] = i+'A';
}
strArray[i] = 0;
printf("Null terminated array of char: %s\n", strArray);
unsigned u = ...;
if (0x10 > u)
exit(EXIT_FAILURE);
while (0x10000 < u) u /= 2;
while (0x1000 > u) u *= 2;
char c[2] = {u / 0x100, u % 0x100);