请问long型怎么转化为char*？

[sdyqingdao]

如题

[al0n9]

long longVar = 0x30;
char charVar = (char)long;

char strBuf[20];

sprintf(strBuf, "%d", longVar);

[blackmurder]

楼上正解

[expter]

long l;
char a[10];
ltoa(l,a,10);

[Mr-Chen]

sprintf

[lidongri]

long l;
char a[10];
ltoa(l,a,10);

[jenycheng]

char szVal[MAX_PATH] = {0} ;
long lVal = 0x30 ;

ltoa (szVal, lVal, 10) ;

[zqche]

_ltoa(l, a, x)

[yoyo_alex_lw]

char *_ltoa(
long value,
char *str,
int radix
);
wchar_t *_ltow(
long value,
wchar_t *str,
int radix
);
template <size_t size>
char *_ltoa(
long value,
char (&str)[size],
int radix
); // C++ only
template <size_t size>
wchar_t *_ltow(
long value,
wchar_t (&str)[size],
int radix
); // C++ only
详细参考msdn.请养成随时查找msdn的习惯

[amei_zhang83]

哈哈，路过，学习下，我就知道这个转换！_ltoa(l, a, x)

[gjw310]

字符串类的format函数也可以