[Bug] LaTeX 渲染异常

[d0j1a1701]

Bug Description

复现方法：在输入框中输入以下内容

在 C 和 C++ 语言中，对于无符号整数类型（如 `unsigned int`），表达式 `(unsigned int)-1` 并不是未定义行为（Undefined Behavior, UB）。这是因为当你对一个无符号类型进行赋值时，如果赋值的数值超出了该类型能表示的范围，结果会是该数值对无符号类型所能表示的最大数值加一的模（即数值在该类型范围内“环绕”）。

对于 `(unsigned int)-1`，由于 `-1` 首先被视为一个有符号整数，然后它被转换（或称为“强制转换”）为 `unsigned int` 类型。在这种情况下，`-1` 转换为无符号类型的方式是通过加上 `unsigned int` 能表示的最大数值加一（即加上 $2^{32}-1$，如果 `unsigned int` 是 32 位的话）。结果就是 `unsigned int` 能表示的最大值，通常是 $2^{32}-1$ 对于 32 位的 `unsigned int`。

因此，这种转换是定义良好的，而不是未定义行为。这一行为在 C 和 C++ 标准中明确指出，因此你可以依赖这种行为来进行编程。

它应当被渲染为：

在 C 和 C++ 语言中，对于无符号整数类型（如 unsigned int），表达式 (unsigned int)-1 并不是未定义行为（Undefined Behavior, UB）。这是因为当你对一个无符号类型进行赋值时，如果赋值的数值超出了该类型能表示的范围，结果会是该数值对无符号类型所能表示的最大数值加一的模（即数值在该类型范围内“环绕”）。

对于 (unsigned int)-1，由于 -1 首先被视为一个有符号整数，然后它被转换（或称为“强制转换”）为 unsigned int 类型。在这种情况下，-1 转换为无符号类型的方式是通过加上 unsigned int 能表示的最大数值加一（即加上 $2^{32}-1$，如果 unsigned int 是 32 位的话）。结果就是 unsigned int 能表示的最大值，通常是 $2^{32}-1$ 对于 32 位的 unsigned int。

因此，这种转换是定义良好的，而不是未定义行为。这一行为在 C 和 C++ 标准中明确指出，因此你可以依赖这种行为来进行编程。

实际却被渲染为：

Steps to Reproduce

复现方法：在输入框中输入以下内容

在 C 和 C++ 语言中，对于无符号整数类型（如 `unsigned int`），表达式 `(unsigned int)-1` 并不是未定义行为（Undefined Behavior, UB）。这是因为当你对一个无符号类型进行赋值时，如果赋值的数值超出了该类型能表示的范围，结果会是该数值对无符号类型所能表示的最大数值加一的模（即数值在该类型范围内“环绕”）。

对于 `(unsigned int)-1`，由于 `-1` 首先被视为一个有符号整数，然后它被转换（或称为“强制转换”）为 `unsigned int` 类型。在这种情况下，`-1` 转换为无符号类型的方式是通过加上 `unsigned int` 能表示的最大数值加一（即加上 $2^{32}-1$，如果 `unsigned int` 是 32 位的话）。结果就是 `unsigned int` 能表示的最大值，通常是 $2^{32}-1$ 对于 32 位的 `unsigned int`。

因此，这种转换是定义良好的，而不是未定义行为。这一行为在 C 和 C++ 标准中明确指出，因此你可以依赖这种行为来进行编程。

Expected Behavior

它应当被渲染为：

在 C 和 C++ 语言中，对于无符号整数类型（如 unsigned int），表达式 (unsigned int)-1 并不是未定义行为（Undefined Behavior, UB）。这是因为当你对一个无符号类型进行赋值时，如果赋值的数值超出了该类型能表示的范围，结果会是该数值对无符号类型所能表示的最大数值加一的模（即数值在该类型范围内“环绕”）。

对于 (unsigned int)-1，由于 -1 首先被视为一个有符号整数，然后它被转换（或称为“强制转换”）为 unsigned int 类型。在这种情况下，-1 转换为无符号类型的方式是通过加上 unsigned int 能表示的最大数值加一（即加上 $2^{32}-1$，如果 unsigned int 是 32 位的话）。结果就是 unsigned int 能表示的最大值，通常是 $2^{32}-1$ 对于 32 位的 unsigned int。

因此，这种转换是定义良好的，而不是未定义行为。这一行为在 C 和 C++ 标准中明确指出，因此你可以依赖这种行为来进行编程。

Screenshots

Deployment Method

• Docker
• Vercel
• Server

Desktop OS

Windows 11

Desktop Browser

Microsoft Edge

Desktop Browser Version

123

Smartphone Device

No response

Smartphone OS

No response

Smartphone Browser

No response

Smartphone Browser Version

No response

Additional Logs

No response

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Title: [Bug] LaTeX rendering exception

Bug Description

How to reproduce: Enter the following content in the input box

In the C and C++ languages, the expression `(unsigned int)-1` is not Undefined Behavior (UB) for unsigned integer types such as `unsigned int`. This is because when you assign a value to an unsigned type, if the assigned value exceeds the range that the type can represent, the result will be the value modulo the maximum value that the unsigned type can represent plus one (that is, the value is within "surround" within the scope of this type).

In the case of `(unsigned int)-1`, since `-1` is first treated as a signed integer, then it is converted (or "cast") to the `unsigned int` type. In this case, `-1` is converted to an unsigned type by adding the largest value that can be represented by `unsigned int` plus one (that is, adding $2^{32}-1$, if `unsigned int` If it is 32 bits). The result is the maximum value that an `unsigned int` can represent, which is typically $2^{32}-1$ for a 32-bit `unsigned int`.

Therefore, this conversion is well-defined rather than undefined behavior. This behavior is explicit in the C and C++ standards, so you can rely on this behavior for your programming.

It should be rendered as:

In the C and C++ languages, the expression (unsigned int)-1 is not Undefined Behavior (UB) for unsigned integer types such as unsigned int. This is because when you assign a value to an unsigned type, if the assigned value exceeds the range that the type can represent, the result will be the value modulo the maximum value that the unsigned type can represent plus one (that is, the value is within "surround" within the scope of this type).

In the case of (unsigned int)-1, since -1 is first treated as a signed integer, then it is converted (or "cast") to the unsigned int type. In this case, -1 is converted to an unsigned type by adding the largest value that can be represented by unsigned int plus one (that is, adding $2^{32}-1$, if unsigned int If it is 32 bits). The result is the maximum value that an unsigned int can represent, which is typically $2^{32}-1$ for a 32-bit unsigned int.

Therefore, this conversion is well-defined rather than undefined behavior. This behavior is explicit in the C and C++ standards, so you can rely on this behavior for your programming.

It is actually rendered as:

Steps to Reproduce

How to reproduce: Enter the following content in the input box

In the C and C++ languages, the expression `(unsigned int)-1` is not Undefined Behavior (UB) for unsigned integer types such as `unsigned int`. This is because when you assign a value to an unsigned type, if the assigned value exceeds the range that the type can represent, the result will be the value modulo the maximum value that the unsigned type can represent plus one (that is, the value is within "surround" within the scope of this type).

In the case of `(unsigned int)-1`, since `-1` is first treated as a signed integer, then it is converted (or "cast") to the `unsigned int` type. In this case, `-1` is converted to an unsigned type by adding the largest value that can be represented by `unsigned int` plus one (that is, adding $2^{32}-1$, if `unsigned int` If it is 32 bits). The result is the maximum value that an `unsigned int` can represent, which is typically $2^{32}-1$ for a 32-bit `unsigned int`.

Therefore, this conversion is well-defined rather than undefined behavior. This behavior is explicit in the C and C++ standards, so you can rely on this behavior for your programming.

Expected Behavior

It should be rendered as:

In the C and C++ languages, the expression (unsigned int)-1 is not Undefined Behavior (UB) for unsigned integer types such as unsigned int. This is because when you assign a value to an unsigned type, if the assigned value exceeds the range that the type can represent, the result will be the value modulo the maximum value that the unsigned type can represent plus one (that is, the value is within "surround" within the scope of this type).

In the case of (unsigned int)-1, since -1 is first treated as a signed integer, then it is converted (or "cast") to the unsigned int type. In this case, -1 is converted to an unsigned type by adding the largest value that can be represented by unsigned int plus one (that is, adding $2^{32}-1$, if unsigned int If it is 32 bits). The result is the maximum value that an unsigned int can represent, which is typically $2^{32}-1$ for a 32-bit unsigned int.

Therefore, this conversion is well-defined rather than undefined behavior. This behavior is explicit in the C and C++ standards, so you can rely on this behavior for your programming.

Screenshots

Deployment Method

• Docker
• Vercel
• Server

Desktop OS

Windows 11

Desktop Browser

Microsoft Edge

Desktop Browser Version

123

Smartphone Device

No response

Smartphone OS

No response

Smartphone Browser

No response

Smartphone Browser Version

No response

Additional Logs

No response

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This bug related to this issue

• [Bug] LaTeX Syntax still bug #3239